Show that $(a^2-b^2)(a^4-b^4) \leq (a^3-b^3)^2$ and $(a^2+b^2)(a^4+b^4)\geq(a^3+b^3)^2$ for all $a,b$.

I'm basically new to inequality. I'm starting my first inequality book which is Introduction to Inequality

So this is the first question I stuck on. Hope someone could provide some hints for it. Thanks in advance.

  • $\begingroup$ Hint: expand out both and eliminate common terms. You should find that both come down to the same inequality; divide out by a common factor and you should find a pretty simple AGM instance... $\endgroup$ – Steven Stadnicki May 28 '17 at 7:38

Let's look at the first one. It's a good idea to take the difference of the two sides, viz., $$ \begin{align} &(a^3-b^3)^2-(a^2-b^2)(a^4-b^4)\\ &=-2a^3b^3+a^4b^2+a^2b^4\\ &=a^2b^2(a^2-2ab+b^2). \end{align} $$ Can you see why this expression is $\ge0$?

  • $\begingroup$ So it means $a^2b^2(a-b)^2 \geq 0$. Am I right> $\endgroup$ – Mathxx May 28 '17 at 7:42
  • $\begingroup$ @Mathxx Yes, indeed you are. $\endgroup$ – Lord Shark the Unknown May 28 '17 at 7:46
  • $\begingroup$ Yea. I solved both inequalities. Thanks a lot! I'm so stupid that I didn't notice this trick! $\endgroup$ – Mathxx May 28 '17 at 7:47

For the second one.

By Cauchy- Schwarz, $(a^2+b^2)((a^2)^2+(b^2)^2) \geqslant (a \cdot a^2+b \cdot b^3)^2=(a^3+b^3)^2$

  • $\begingroup$ Will come back for it when I've learnt Cauchy-Schwarz ! $\endgroup$ – Mathxx May 28 '17 at 7:49

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