<s>[INST] General Inequality Problem. The question goes like this :

Let $a,b,c,d$  be positive reals  and  given that  $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$

My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although I am not sure if its the correct method, I tried reducing the powers.
Then, I got:
$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $
I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!
 [/INST] By Chebyshev's inequality,
$\frac{a^2+b^2+c^2+d^2}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a+b+c+d}{4}$
$\implies a^2+b^2+c^2+d^2 \geq \frac{1}{4}$
Similarly,
$\frac{a^3+b^3+c^3+d^3}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a^2+b^2+c^2+d^2}{4}$
$\implies 4(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2$
$6(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2+ \frac{a^2+b^2+c^2+d^2}{2} \geq a^2+b^2+c^2+d^2+ \frac{1}{8}$
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