Show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
$begingroup$
How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
Assuming that :
$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
So
$(sqrt{x-1}+sqrt{y-1})^2leq xy$
$sqrt{(x-1)(y-1)} leq xy-x-y+2$
$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$
Here I'm stuck !
algebra-precalculus inequality radicals
asked Nov 8 '15 at 21:10
user233658user233658
66649
$endgroup$
add a comment |
$begingroup$
How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
Assuming that :
$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
So
$(sqrt{x-1}+sqrt{y-1})^2leq xy$
$sqrt{(x-1)(y-1)} leq xy-x-y+2$
$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$
Here I'm stuck !
algebra-precalculus inequality radicals
asked Nov 8 '15 at 21:10
user233658user233658
66649
$endgroup$
$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13
$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25
add a comment |
$begingroup$
How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
Assuming that :
$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
So
$(sqrt{x-1}+sqrt{y-1})^2leq xy$
$sqrt{(x-1)(y-1)} leq xy-x-y+2$
$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$
Here I'm stuck !
algebra-precalculus inequality radicals
asked Nov 8 '15 at 21:10
user233658user233658
66649
$endgroup$
How can one show that $sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
Assuming that :
$sqrt{x-1}+sqrt{y-1}leq sqrt{xy}$
So
$(sqrt{x-1}+sqrt{y-1})^2leq xy$
$sqrt{(x-1)(y-1)} leq xy-x-y+2$
$ (y-1)(x-1)+3 leq sqrt{(x-1)(y-1)}$
Here I'm stuck !
algebra-precalculus inequality radicals
algebra-precalculus inequality radicals
asked Nov 8 '15 at 21:10
user233658user233658
66649
asked Nov 8 '15 at 21:10
user233658user233658
66649
edited Nov 8 '15 at 21:25
user147263
asked Nov 8 '15 at 21:10
user233658user233658
66649
asked Nov 8 '15 at 21:10
user233658user233658
66649
asked Nov 8 '15 at 21:10
user233658user233658
66649
66649
$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13
$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25
add a comment |
$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13
$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25
$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13
$begingroup$
$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
$endgroup$
– Ross Millikan
Nov 8 '15 at 21:13
$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25
$begingroup$
Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
$endgroup$
– user147263
Nov 8 '15 at 21:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
add a comment |
$begingroup$
Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$
$sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
add a comment |
$begingroup$
There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
$endgroup$
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
add a comment |
$begingroup$
There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
$endgroup$
There are some errors in your calculation, e.g. a missing factor 2 in
$$
(sqrt{x-1}+sqrt{y-1})^2 = x - 1 + y - 1 + 2sqrt{x-1}sqrt{y-1}
$$
and in the last step the inequality sign is in the wrong direction and
the number $3$ is wrong.
For $x ge 1$, $y ge 1$ you can square the inequality (since both
sides are non-negative):
$$
sqrt{mathstrut x-1}+sqrt{mathstrut y-1}leq sqrt{mathstrut xy} \
Longleftrightarrow (x-1) + (y-1) + 2 sqrt{mathstrut x-1}sqrt{mathstrut y-1} le xy \
Longleftrightarrow 0 le xy - x - y + 2 - 2 sqrt{mathstrut (x-1)(y-1)} \
Longleftrightarrow 0 le (x-1)(y-1) - 2 sqrt{mathstrut (x-1)(y-1)} + 1
$$
With $t := sqrt{(x-1)(y-1)}$ the right-hand side is
$$
t^2 - 2 t + 1 = (t-1)^2 ge 0 , .
$$
so that the inequality is true.
It follows also that equality holds if and only if $t = 1$,
i.e. if $(x-1)(y-1) = 1$.
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
edited Nov 8 '15 at 22:30
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
answered Nov 8 '15 at 21:23
Martin RMartin R
27.8k33255
27.8k33255
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
add a comment |
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
Thank you so much for the answer ! :)
$endgroup$
– user233658
Nov 8 '15 at 21:38
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
$begingroup$
@user233658: You are welcome!
$endgroup$
– Martin R
Nov 8 '15 at 21:43
add a comment |
$begingroup$
Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
$endgroup$
Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=sqrt{x-1},quad u=sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+ulesqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+uge0$, the inequality is equivalent to
$$
(t+u)^2le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
answered Nov 8 '15 at 22:01
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$
$sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$
$sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$
$sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
WLOG $sqrt x=sec A,sqrt y=sec B$ where $0<A,B<dfracpi2$
$sqrt{x-1}+sqrt{y-1}=tan A+tan B=dfrac{sin(A+B)}{cos Acos B}lesqrt{sec Asec B}=?$
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
answered Jan 1 at 4:03
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
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$LaTeX$ hint: If you put braces around everything that belongs under the square root sign, the top bar extends to cover it. Putting sqrt {x-1} between dollar signs gives $sqrt {x-1}$
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– Ross Millikan
Nov 8 '15 at 21:13
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Assuming what needs to be proved is never a good idea. (A valid method of proof is proof by contradiction, which would begin by assuming the opposite of what needs to be proved, and reach a contradiction from there.)
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– user147263
Nov 8 '15 at 21:25