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$a$ and $b$ are real non-negative numbers, for which $a^2 + b^2 = 4$. What is the easiest way to prove the following inequality (without using any fancy inequalities like Jensen's or Muirhead's)?

$\frac{ab}{a + b + 2} \le \sqrt{2} - 1$

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2 Answers 2

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Since $$((a+b)-2)((a+b)+2)=(a^2+b^2+2ab-4)=2ab,$$ we have $$\frac{ab}{a+b+2}=\frac{a+b-2}{2}=\frac{a+b}{2}-1.$$ Note that it is a circle and question is related with the maximum value of $$2\cos\theta+2\sin\theta$$ which is $2\sqrt 2$.

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  • $\begingroup$ Great answer! $+1$ $\endgroup$ Apr 18, 2016 at 19:03
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You may use the fact that $a^2+b^2=4$ is the equation of a circle while $ab=(\sqrt{2}-1)(a+b+2)$ is the equation of a rectangular hyperbola, and both conic sections go through the point $(\sqrt{2},\sqrt{2})$, but with different curvatures. In particular, the line with equation $a+b=2\sqrt{2}$ is tangent to both curves, and $$a^2+b^2=4,\quad a,b\geq 0$$ implies $a+b\leq 2\sqrt{2}$, that implies $ab\leq(\sqrt{2}-1)(a+b+2)$.

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