Prove that $-(p_1+p_2)\log{p_1+p_2} \leq -p_1 \log{p_1} - p_2 \log{p_2}$ provided that $ p_1,p_2 > 0$ WTS:
$$-(p_1+p_2)\log{(p_1+p_2)} \leq -p_1 \log{p_1} - p_2 \log{p_2} \> \> \forall \> \> p_1,p_2 > 0$$
Any hints on this? I've tried to set it up as a proof by contradiction, and jiggled around with the terms both as logarithms and after exponentiating everything, but no luck. 
 A: $p_1\log(p_1) + p_2\log(p_2) - p_1\log(p_1+p_2) - p_2\log(p_1+p_2) =p_1(\log(p_1)-\log(p_1+p_2)) +p_2(\log(p_2)-\log(p_1+p_2))\leq 0 + 0 = 0$
A: We have to prove:
$$
f(x,y)=x\ln x+y \ln y- (x+y)\ln(x+y)\leq 0
$$
Since for a given $y>0$:
$$
\frac{\partial f}{\partial x}(x,y)=\ln x-\ln(x+y)=\ln\left(\frac{x}{x+y}\right)\leq 0
$$
And equality occurs when $x=0$.
$$
\lim_{x\to 0}f(x,y)=0
$$
We are done by symmetry, and $f(0,0)=0$.
Or version $B:$ Notice that for $y>0$:
$$
\int_{0}^x 0\mathrm{d}t\geq \int_{0}^y\ln\left(\frac{t}{t+x}\right)\mathrm{d}t=
$$
$$
=\left.-t \log (t+x)-x \log (t+x)+t \log (t)\right|_0^y=(x+y)\ln(x+y)-x\ln x-y\ln y
$$
A: $$-(p_1+p_2)\log{(p_1+p_2)} \leq -p_1 \log{p_1} - p_2 \log{p_2}$$
$$\Leftrightarrow (p_1+p_2)\log{(p_1+p_2)} \geq p_1 \log{p_1} + p_2 \log{p_2}$$
$$\Leftrightarrow log{(p_1+p_2)^{p_1+p_2}} \geq \log{p_1^{p_1}} + \log{p_2^{p_2}}$$
$$\Leftrightarrow log{(p_1+p_2)^{p_1+p_2}} \geq \log{(p_1^{p_1}\times p_2^{p_2})}$$
$$\Leftrightarrow (p_1+p_2)^{p_1+p_2} \geq p_1^{p_1}\times p_2^{p_2}$$
$$\Leftrightarrow \frac{(p_1+p_2)^{p_1}(p_1+p_2)^{p_2}}{p_1^{p_1}\times p_2^{p_2}}\geq1$$
$$\Leftrightarrow \frac{(p_1+p_2)^{p_1}}{p_1^{p_1}}\times\frac{(p_1+p_2)^{p_2}}{p_2^{p_2}}\geq1$$
$$\Leftrightarrow \left(1+\frac{p_2}{p_1}\right)^{p_1}\times\left(1+\frac{p_1}{p_2}\right)^{p_2}\geq1$$
