# 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.

## 1 Answer

It's just C-S: $$3(1+a^2+a^4)=(1+1+1)(1+a^2+a^4)\geq(1+a+a^2)^2.$$

The Chebyshov inequality it's the following.

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.

There is also the following way:

$$(1+a^2+a^4)=(1+2a^2+a^4-a^2)=((1+a^2)^2-a^2)=(1-a+a^2)(1+a+a^2).$$ Id est, it's enough to prove that: $$3(1-a+a^2)\geq1+a+a^2$$ or $$(a-1)^2\geq0.$$