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Let $a,b,c$ be real numbers such that $a+b+c = 0$ and $a^2 + b^2 + c^2 = 1$. Prove that $$a^2b^2c^2\le \frac1{54}$$

Jimmy R.
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Sarnavo Sarkar
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1 Answers1

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By calculation $$ab+bc+ca=-1/2$$ and

$$1/4=(ab+bc+ca)^2=2abc(a+b+c)+a^2 b^2+b^2 c^2+c^2 a^2=a^2 b^2+b^2 c^2+c^2 a^2$$.

Let $K=a^2b^2c^2$, then $a^2,b^2,c^2$ are the solutions of

$$t^3-t^2+\frac{1}{4}t-K=0$$

Thus

$$K=t^3-t^2+\frac{1}{4}t$$

By taking derivative, we know the maximal value $K=1/54$ is achieved at $t=1/6$.

AlgRev
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