# chebyshev's inequality - Question

I had a question in my exam and they asked to prove that prove that:

$$3(1+a^2+a^4)\geq(1+a+a^2)^2$$ for all $$a\in\mathbb R$$.

Now , I solved it , but the problem is that in the answer they wrote this: using Chebyshev inequality:

$$(1+a+a^2)^2=(1·1+a·1+a^2·1)^2≤(1+a^2+a^4)·(1+1+1)=3(1+a^2+a^4).$$ And so I tried searching the web for this inequality but all it found was the Chebyshev's inequality for probabillity. can someone please send me link regarding this inequality or just write it here?

Thank you.

It's just C-S: $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)\geq(1+a+a^2)^2.$$
Let $$a_1\geq a_2\geq...\geq a_n$$ and $$b_1\geq b_2\geq...\geq b_n$$.
Thus, $$n\sum_{k=1}^na_kb_{n-k+1}\leq\sum_{k=1}^na_k\sum_{k=1}^nb_k\leq n\sum_{k=1}^na_kb_k.$$ The proof of this inequality follows immediately from Rearrangement.
$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$ Id est, it's enough to prove that: $$3(1-a+a^2)\geq1+a+a^2$$ or $$(a-1)^2\geq0.$$