Dtyjlui

let reals $p, q >1$ and $x, y$ are positive real numbers with $1/p+1/q=1$ I want to show $xyleq frac{1}{p}x^p+frac{1}{q}y^q.$ without using concavity of $log$ as shown below

Proof with concavity of log function and exponential Growth

$lnleft(frac{1}{p}x^p+frac{1}{q}y^qright)geq frac{1}{p}lnleft(x^pright)+frac{1}{q}lnleft(y^qright)=lnleft(xyright).$

just to raise expon for both side we get the result .

then any other method ?

Method 1:

For non-negative $u,v$ let $f(u)=u^{p-1}$ and $g(v)=v^{q-1}.$ Since $(p-1)(q-1)=1,$ the graphs of $(u,f(u))$ and of $(g(v),v)$ are the same graph, which I will call $G.$

Now $x^p/p=int_0^xf(u)du$ is the area of the figure bounded by $G$ and ${x}times [0,f(x)]$ and the $u$-axis in the $(u,v)$-plane. And $y^q/q=int_0^yg(y)dy$ is the area of the figure bounded by $G$ and $[0, g(y)]times {y}$ and the $v$-axis.

If $x^p=y^q,$ the union of these two figures is the rectangle $R(x,y)$ with corners $(0,0),(0,y),(x,y),(x,0),$ which has area $xy,$ so $x^p/p+y^q/q=int_0^xf(u)du+int_0^yg(v)dv=xy.$

If $x>y^{q/p}$ then the union of these two figures is $R(x,y)$ plus a region bounded by $G$, the $u$-axis, and the segments ${y^{q/p}}times [0,y]$ and ${x}times [0,f(x)],$ so $x^p/p+y^q/q=xy+int_{y^{q/p}}^x f(u)du>xy$.

Similarly if $x<y^{p/q}$ then $y>x^{q/p}$ and $x^p/p+y^q/q=xy+int_{x^{q/p}}^y g(v)dv>xy.$

This is easy to see on a diagram.

Generally, if $f:[0,infty)to [0,infty)$ is continuous and strictly monotonic, and $f(0)=0,$ then $xyleq int_0^xf(u)du+int_0^yf^{-1}(v)dv.$

Method 2:

For non-negative $x,y$ let $H(x,y)=x^p/p+y^q/q-xy.$ Then $frac {partial H(x,y)}{partial x}=frac1px^{p-1}-y,$ which is negative if $0leq x^{p-1}<y$ and positive if $x^{p-1}>y.$ So for a fixed $y,$ the function $H(x,y)$ has its minimum when $x^{p-1}=y.$ And we have $H(x, x^{p-1})=0.$

This proof is from Steele's Cauchy-Schwarz Master Class. The weighted AM-GM inequality gives us
begin{align*}
u^alpha v^beta le frac{alpha}{alpha + beta}u^{alpha + beta} + frac{beta}{alpha + beta}v^{alpha + beta}
end{align*}
for $u, v ge 0$ and $alpha, beta > 0$. Substituting $x = u^alpha$, $y= v^beta$, $p = (alpha+beta)/alpha$, and $q = (alpha + beta)/beta$ gives us the desired inequality.

Another cool proof is using Legendre's transform, with $L(x) = frac{1}{p}|x|^p$ is convex, then
$$ L^*(y) = sup_{xin mathbb{R}} Big(xcdot y - L(y)Big) = frac{1}{q}|y|^q.$$
Therefore
$$ L(x)+L^*(y) = frac{1}{p}|x|^p + frac{1}{q}|y|^q geq xcdot{y}.$$

More on this, see https://en.wikipedia.org/wiki/Legendre_transformation. It is also called "Fenchel's inequality".

Hint

Let $$f(t)=frac{t^p}{p}+frac{1}{q}-t text{ on } [0,infty)$$
Now show that $t=1$ is a critical point and $f$ is an increasing function for $t geq 1$. This will imply
$$f(t) geq f(1)=0.$$
Consequently,
$$frac{t^p}{p}+frac{1}{q} geq t$$
Now let $t=xy^{frac{1}{p-1}}$.