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Question Number 115815 by Ar Brandon last updated on 28/Sep/20

Montrer que \forall (a, b, c) \in {(R_{+}^{*})}^{3}

\frac{1}{a^{2} + bc} + \frac{1}{b^{2} + ac} + \frac{1}{c^{2} + ab} ⩽ \frac{1}{2} (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac})

Answered by 1549442205PVT last updated on 29/Sep/20

We have :

\frac{1}{a^{2} + bc} + \frac{1}{b^{2} + ac} + \frac{1}{c^{2} + ab} ⩽ \frac{1}{2} (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac})

\Leftrightarrow \frac{1}{a^{2} + bc} + \frac{1}{b^{2} + ac} + \frac{1}{c^{2} + ab} ⩽ \frac{a + b + c}{2 abc} (1)

Applying {Cauchy}^{'} s inequality for two

positive numbers we have :

a^{2} + bc ⩾ 2 a \sqrt{bc}, b^{2} + ac ⩾ 2 b \sqrt{ac}

c^{2} + ab ⩾ 2 \sqrt{ab}, it follows that

L . H . S ⩽ \frac{1}{2 a \sqrt{bc}} + \frac{1}{2 b \sqrt{ac}} + \frac{1}{2 c \sqrt{ab}}

Hence, we just need prove that :

\frac{1}{2 a \sqrt{bc}} + \frac{1}{2 b \sqrt{ac}} + \frac{1}{2 c \sqrt{ab}} ⩽ \frac{a + b + c}{2 abc} (2)

\Leftrightarrow \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} ⩽ \frac{a + b + c}{\sqrt{abc}}

\Leftrightarrow \sqrt{ab} + \sqrt{bc} + \sqrt{ca} ⩽ a + b + c

\Leftrightarrow 2 (\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) ⩽ 2 (a + b + c)

\Leftrightarrow {(\sqrt{a} - \sqrt{b})}^{2} + {(\sqrt{b} - \sqrt{c})}^{2} + {(\sqrt{c} - \sqrt{a})}^{2} ⩾ 0

This inequality is always true

\forall a, b, c > 0 . Hence, (2) is true infer (1) is

true . Thus, the given inequality is

proved . The equality ocurrs if and only

if a = b = c . (Q . E . D)