Can anyone offer some guidance on proving the following inequality?
For any $a,b,c,d,e \geq 0$ ,$p > 1$ and $ \varepsilon > 0$, the following holds: $$(a+b+c+d+e)^p \leq 2^p(1+\frac{1}{\varepsilon})(b^p+c^p+d^p+e^p)+(1+\varepsilon)^{\frac{p}{2}} a^p.$$
To start with this，I want to prove the simple case:$$(a+b)^p \leq 2^p(1+\frac{1}{\varepsilon})b^p+(1+\varepsilon)^{\frac{p}{2}} a^p.$$ First， if $a \;\text{or} \;b =0$, then it is obvious. Hence the inequality is equivalent to $$(\frac{1+a}{2})^p \leq (1+\frac{1}{\varepsilon})a^p+(\sqrt{\frac{1+\varepsilon}{4}})^p,$$ where $a > 0$. Due to the convexity，the simple case is proved.
However, I can't seem to progress further from this.