Inequalities help: $(a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)$

Question

Prove: that $$(a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)$$ for all real numbers $a$ and $b$.

My attempt:

$(a^2+b^2)(a^4+b^4)=a^6+a^2 b^4+b^2 a^4+b^6\geq 4\cdot(a^6\cdot b^6\cdot a^2 b^4\cdot a^4 b^2)^{\frac{1}{4}}=4\cdot a^3\cdot b^3$, by AM-GM.

Now $(a^3+b^3)^2\geq 4a^3\cdot b^3$, by AM-GM. putting this above gives:$$(a^2+b^2)(a^4+b^4)\geq 4\cdot a^3\cdot b^3\leq (a^3+b^3)^2$$

But this don't gives good, please help, on what I did. How to improve solving Inequalities?

Answer 1

$$(a^2+b^2)(a^4+b^4)-(a^3+b^3)^2=(ab^2-a^2b)^2$$

Answer 2

we have $$(a^2+b^2)(a^4+b^4)-(a^3+b^3)^2=a^2b^2(a-b)^2\geq 0$$

Answer 3

You have the reversible steps $$a^6+2a^3b^3+b^6\le a^6+a^4b^2+a^2b^4+b^6$$

$$2a^3b^3\le a^4b^2+a^2b^4$$

Now note that $ab=0$ gives cases which make the original inequality trivially true, so we can assume $a^2b^2\gt 0$ and divide by $2a^2b^2$ $$ab\le \frac {a^2+b^2}2$$ which is just the AM/GM inequality applied to $a^2$ and $b^2$ - but just as easy to observe that $(a-b)^2\ge 0$ (which is an easy way to prove AM/GM for two numbers).

Answer 4

By Cauchy-Schwarz $$(a^4+b^4)(a^2+b^2)\geq\left(a^2\cdot a+b^2\cdot b\right)^2=(a^3+b^3)^2$$