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I was asked to prove that

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

I expanded the rhs and used am gm and got $a^6+a^2b^4+a^4b^2+b^6 \ge 4a^3b^3$ and struck.i think I lack intuition or I don't have enough experience.Any hints

Mathematical Curiosity
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  • This inequality is false as stated: take $a=1/2$ and $b=0$. What values are $a$ and $b$ allowed to take? Should the left-hand side be squared? If so, it's the Cauchy-Schwarz inequality, which can be interpreted here as saying that the square of the dot product of the vectors $(a,b)$ and $(a^2,b^2)$ is bounded above by the product of their norms. – Greg Martin Mar 17 '20 at 04:41
  • @ Greg Martin maybe my textbook has a typo.Anyways thank you for your reaponse – Mathematical Curiosity Mar 17 '20 at 05:02

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Maybe you want to prove: $(a^3+b^3)^2 \le (a^2+b^2)(a^4+b^4)$. In this case, it reduces to: $2a^3b^3 \le a^2b^4+a^4b^2$, but this is true because $a^2b^4+a^4b^2 - 2a^3b^3 = (ab^2-a^2b)^2 \ge 0$ clearly true.

DeepSea
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