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:

$$-(p_1+p_2)\log{(p_1+p_2)} \leq -p_1 \log{p_1} - p_2 \log{p_2} \&gt; \&gt; \forall \&gt; \&gt; p_1,p_2 &gt; 0$$

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.
$p_1\log(p_1) + p_2\log(p_2) - p_1\log(p_1+p_2) - p_2\log(p_1+p_2) =p_1(\log(p_1)-\log(p_1+p_2)) +p_2(\log(p_2)-\log(p_1+p_2))\leq 0 + 0 = 0$
$$-(p_1+p_2)\log{(p_1+p_2)} \leq -p_1 \log{p_1} - p_2 \log{p_2}$$
$$\Leftrightarrow (p_1+p_2)\log{(p_1+p_2)} \geq p_1 \log{p_1} + p_2 \log{p_2}$$
$$\Leftrightarrow log{(p_1+p_2)^{p_1+p_2}} \geq \log{p_1^{p_1}} + \log{p_2^{p_2}}$$
$$\Leftrightarrow log{(p_1+p_2)^{p_1+p_2}} \geq \log{(p_1^{p_1}\times p_2^{p_2})}$$
$$\Leftrightarrow (p_1+p_2)^{p_1+p_2} \geq p_1^{p_1}\times p_2^{p_2}$$
$$\Leftrightarrow \frac{(p_1+p_2)^{p_1}(p_1+p_2)^{p_2}}{p_1^{p_1}\times p_2^{p_2}}\geq1$$
$$\Leftrightarrow \frac{(p_1+p_2)^{p_1}}{p_1^{p_1}}\times\frac{(p_1+p_2)^{p_2}}{p_2^{p_2}}\geq1$$
$$\Leftrightarrow \left(1+\frac{p_2}{p_1}\right)^{p_1}\times\left(1+\frac{p_1}{p_2}\right)^{p_2}\geq1$$