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Question: If $a,b\in \mathbb R^+$, $|a-2b|\leq\frac {1}{\sqrt{a}}$, $|b-2a|\leq\frac {1}{\sqrt{b}},$ prove $a+b≤2$.

I figured out that $a+b\leq \frac {1}{\sqrt{a}}+\frac {1}{\sqrt{b}}$, but I am not sure how to prove that $a+b\leq 2$ after doing this.

Can anyone help me?

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  • $\begingroup$ Hint: Note that the limiting case occurs when $a = b = 1$. Also, think about the magnitudes involved in the inequality. Is there something you can use there? $\endgroup$
    – user0
    Mar 6, 2021 at 16:54

2 Answers 2

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Note by given condition$$a{(a-2b)}^2+b{(b-2a)}^2\le 1+1 $$ $$\iff a^3+b^3\le 2$$ Now use by power-mean inequality that $$\frac{{(a+b)}^3}{4}\le a^3+b^3$$

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    $\begingroup$ @ Albus Dumbledore: Your proof implies $a+b \le 2\sqrt[3]{2}$, but it said in the post that $a+b \le 2$. So you are still a bit less what is required... $\endgroup$
    – user887708
    Mar 6, 2021 at 17:46
  • $\begingroup$ @DalatKarelia it was a typo please see now $\endgroup$
    – Albus Dumbledore
    Mar 6, 2021 at 17:48
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    $\begingroup$ A nice solution $\endgroup$
    – user822157
    Mar 7, 2021 at 9:38
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Just my 2 cents to the Albus' answer, without using power mean. We can assume $0<a\leq b$ ($a=0$ can be checked easily) then $$\frac{(a+b)^2}{4}\leq a^3+b^3 \iff \frac{\left(1+\frac{b}{a}\right)^3}{4}\leq 1+\left(\frac{b}{a}\right)^3 \tag{1}$$ Now $$f(x)=\frac{\left(1+x\right)^3}{4} -1-x^3=(1+x)\left(\frac{(1+x)^2}{4}-(1-x+x^2)\right) =\\ (x+1)\left(\frac{-3+6x-3x^2}{4}\right)=-\frac{3}{4}(x+1)(x-1)^2$$ We can see that $f(x)\leq 0$ for $x\geq1$, which explains $(1)$.

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