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Algebra

Blazing goIITian

Joined:	27 Dec 2010
Post:	1474

11 Feb 2012 08:50:18 IST

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216

good question.

Mathematics

if a and b are real numbers,

such that a > or equal to b > 0.

prove that [(a-b)^2]/[8a] < or equal to (a+b)/2 - sqrt(ab) < or equal to [(a-b)^2]/8b.

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jagdish singh

Blazing goIITian

Joined: 19 Jan 2008

Posts: 805

17 Feb 2012 15:29:26 IST

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$\hspace{-16} \mathbf{\frac{(a-b)^2}{8a}\leq \frac{a+b}{2}-\sqrt{ab}\leq \frac{(a-b)^2}{8b}}\\\\\\ \mathbf{\frac{(a-b)^2}{8a}\leq \frac{(\sqrt{a}-\sqrt{b})^2}{2}\leq \frac{(a-b)^2}{8b}}$\\\\\\ $\mathbf{\frac{(\sqrt{a}-\sqrt{b})^2.(\sqrt{a}+\sqrt{b})^2}{a}\leq 4.(\sqrt{a}-\sqrt{b})^2\leq \frac{(\sqrt{a}-\sqrt{b})^2.(\sqrt{a}+\sqrt{b})^2}{b}}$\\\\\\ $\mathbf{\frac{(\sqrt{a}+\sqrt{b})^2}{a}\leq 4 \leq \frac{(\sqrt{a}+\sqrt{b})^2}{b}}$\\\\\\ $\mathbf{\left(1+\frac{\sqrt{b}}{\sqrt{a}}\right)^2\leq 4 \leq \left(1+\frac{\sqrt{a}}{\sqrt{b}}\right)}$\\\\\\ $\mathbf{1+\frac{\sqrt{b}}{\sqrt{a}}\leq 2 \leq 1+\frac{\sqrt{a}}{\sqrt{b}}}$\\\\\\ $\mathbf{\frac{\sqrt{b}}{\sqrt{a}}\leq 1 \leq \frac{\sqrt{a}}{\sqrt{b}}}$\\\\\\ So $\mathbf{a\geq b>0}$$

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