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

Solution 1:

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

Solution 2:

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

Solution 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).