Apply mean value theorem for inequalities Let $p, q$ be positive real numbers such that $\frac{1}{p} + \frac{1}{q} = 1$. For real numbers $a\geq0, b\geq0$ show that $$ab\leq\frac{a^p}{p}+\frac{b^q}{q}$$
I am newly studying calculus in college. 
I went through the proof present at http://www.math.ust.hk/~majhu/Math203/Rudin/Homework23.pdf 
But I couldn't understand it properly. I am looking for much simpler proof. 
 A: You can start from the inequality:
$$
x^{\lambda} y^{1-\lambda} \leq \lambda x + (1-\lambda) y,
\qquad
x,y>0,\ \lambda\in [0,1],
$$
that can be proved, for example, using the concavity and the monotonicity of the $\log$ function:
$$
\log(\lambda x + (1-\lambda)y) \geq \lambda \log x + (1-\lambda)\log y
= \log(x^\lambda y^{1-\lambda}).
$$
Then you can use this inequality with $x=a^p$, $y=b^q$, $\lambda = 1/p$, $1-\lambda = 1/q$.
A: Another way to show the inequality is to use Young's inequality. 
Young's inequality states that if $f$ is a continuous and strictly increasing function on some interval $[0, c]$ with $c > 0$ and $f(0) = 0$ then for all $a \in [0, c]$ and $b \in [0, f(c)]$
$$\int^a_0 f(x) \, dx + \int^b_0 f^{-1} (x) \, dx \geqslant ab,$$
with equality holding if and only if $b = f(a)$. Here $f^{−1}$ denotes the inverse function of $f$. 
Young’s inequality is so general that many interesting inequalities can
be derived from it. In particular, if the function $f(x) = x^{p-1}$ for $p > 1$ is taken in Young's inequality the result you seek immediately follows.
