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!
 A: 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,
$\frac{a^3+b^3+c^3+d^3}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a^2+b^2+c^2+d^2}{4}$
$\implies 4(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2$
$6(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2+ \frac{a^2+b^2+c^2+d^2}{2} \geq a^2+b^2+c^2+d^2+ \frac{1}{8}$
A: Use Tangent Line Method
$$6x^3-x^2 \geqslant \frac{5x-1}{8}, \quad \forall x > 0.$$
A: Tangent line detailed:
You can rewrite your inequality as $$f(a)+f(b)+f(c)+f(d)\geq {1\over 8}$$ where $f(x)= 6x^3-x^2$. Now we see that we have equality case if $a=b=c=d={1\over 4}$ so it is reasonable to calculate tangent line for $f$ at $x= {1\over 4}$ which is $g(x)={5x-1\over 8}$.
If you draw a graph for $f$ and $g$ you see that $f$ is all the time over $g$ for $x\in[0,1]$, so $$f(x)\geq g(x)$$ This, of course, you should verify with some calculations. So $$f(a)+f(b)+f(c)+f(d)\geq g(a)+g(b)+g(c)+g(d)= {1\over 8}$$
