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Let $a$ and $b$ be positive real numbers. Prove that $$(a+b)^3\le 4(a^3+b^3)$$

My work and thoughts: I've tried the brutal math resulting in: $a^2b+ab^2\le a^3 +b^3$ and not sure where to go from there. Or it can be seen as $a^2b+ab^2\le(a+b)(a^2-ab+b^2).$ Any thoughts, special inequalities, or different methods to help out?

Dave Ezarik
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  • I think my first approach would be taking the derivative of $f(x) = 4(x^3 + b^3) - (x+b)^3$ to see where it reaches a minimum on $[0,+\infty)$. – hardmath Oct 15 '18 at 00:40

5 Answers5

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Hint $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$

By AM-GM $$a^2b \leq \frac{a^3+a^3+b^3}{3} \\ ab^2 \leq \frac{a^3+b^3+b^3}{3} \\$$

N. S.
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You were almost there.

Continuing from where you left off . . . \begin{align*} &a^2b+ab^2\le(a+b)(a^2-ab+b^2)\\[4pt] \iff\;&ab(a+b)\le(a+b)(a^2-ab+b^2)\\[4pt] \iff\;&ab\le a^2-ab+b^2\\[4pt] \iff\;&0\le a^2-2ab+b^2\\[4pt] \iff\;&0\le (a-b)^2\\[4pt] \end{align*}

quasi
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Use Jensen's inequality. The function $f(x)=x^3$ is upwards concave for $x>0$. Therefore

$$({a+b\over 2})^3\leq {a^3+b^3\over 2},$$ $$({a+b})^3\leq 4(a^3+b^3)$$

MrDudulex
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$a^3+b^3 \geq a^2b +ab^2 \Longleftrightarrow a^2(a-b) \geq b^2(a-b) \Longleftrightarrow (a-b)^2(a+b) \geq 0.$

Catalin Zara
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You can also use Holder’s inequality which directly gives: $$(a^3+b^3)^{1/3}(1+1)^{2/3}\geqslant (a+b)$$

Macavity
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