Prove that $p_1\log(\frac{1}{p_1})+p_2\log(\frac{1}{p_2})\leq \log2$ 
If $p_1,p_2>0$ and $p_1+p_2=1$ then prove that $$p_1\log(\frac{1}{p_1})+p_2\log(\frac{1}{p_2})\leq \log2$$

I`ve tried to define $p_1$ as $1-p_2$, and I got $$(\frac{(1-p_2)}{p_2})^{p_2}*\frac{1}{1-p_2}\leq2$$
But I`m not sure what to do next, and I don't know if that's  the right way.
Could you give me any hints? :)
 A: Using the logarithm rules it suffices to show 
$$
\log(p_1^{-p_1}p_2^{-p_2})\leq \log(2),
$$
and since $\log$ is increasing on its domain of definition it suffices to show 
$$
p_1^{-p_1}p_2^{-p_2}\leq 2.
$$
By Young's product inequality (https://en.wikipedia.org/wiki/Young%27s_inequality_for_products) we now have 
$$
p_1^{-p_1}p_2^{-p_2}\leq p_1p_1^{-1}+p_2p_2^{-1}\leq 2.
$$
A: HINT
One simplifying thought is
$$
p \ln \left(\frac1p\right) = p \ln \left(p^{-1}\right) = -p\ln p
$$
so if you let $p \in (0,1)$ with $q=1-p$ you need to show
$$
p \ln(1/p) + q\ln(1/q) \le 2
$$
which is equivalent to
$$
p \ln p + q \ln q \ge -2
$$
and now LHS is
$$
f(p) = p\ln p + (1-p)\ln(1-p)
$$
and you can use Calculus...
UPDATE
To convince you this is indeed the case, here is the graph. The formal proof can be done via 1-variable Calculus.

A: If you are allowed to use the concept of concavity you get immediately from the concavity of $\log x$:


*

*$p\log x + (1-p) \log y \leq \log (px + (1-p)y)$ for all $p \in [0,1]$
It follows
$$p_1\log\left( \frac{1}{p_1} \right)+p_2\log \left(\frac{1}{p_2}\right) \stackrel{p_1+p_2 =1; p_1,p_2 > 0}{\leq} \log \left( \frac{p_1}{p_1} + \frac{p_2}{p_2}\right) = \log2$$
A: Hint

Consider $f(x)=-x\log{x}-(1-x)\log{(1-x)}$. Find the derivative and you’ll get a maximum at some point with value $\log{2}$

A: By tangent line method you can prove $$-x\log x \leq (\log 2-1)x+{1\over 2}$$ is valid for all positive $x$ and thus a conclusion.

The right side is tangent line at point $({1\over 2},\log \sqrt{2})$ on graph of the function $f(x)=-x\log x$.
