let reals $p, q >1$ and $x, y$ are positive real numbers with $1/p+1/q=1$ I want to show $xy\leq \frac{1}{p}x^p+\frac{1}{q}y^q.$ without using concavity of $\log$ as shown below
Proof with concavity of log function and exponential Growth
$\ln\left(\frac{1}{p}x^p+\frac{1}{q}y^q\right)\geq \frac{1}{p}\ln\left(x^p\right)+\frac{1}{q}\ln\left(y^q\right)=\ln\left(xy\right).$
just to raise expon for both side we get the result .
then any other method ?
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.
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 $xy\leq \int_0^xf(u)du+\int_0^yf^{-1}(v)dv.$
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 $0\leq x^{p-1}<y$ 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 $H(x, x^{p-1})=0.$
Another cool proof is using Legendre's transform, with $L(x) = \frac{1}{p}|x|^p$ is convex, then $$ L^*(y) = \sup_{x\in \mathbb{R}} \Big(x\cdot y - L(y)\Big) = \frac{1}{q}|y|^q.$$ Therefore $$ L(x)+L^*(y) = \frac{1}{p}|x|^p + \frac{1}{q}|y|^q \geq x\cdot{y}.$$
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 $t=xy^{\frac{1}{p-1}}$.
