Montrer que ∀(a,b,c)∈(R_+ ^∗ )^3 (1/(a^2 +bc))+(1/(b^2 +ac))+(1/(c^2 +ab))≤(1/2)((1/(ab))+(1/(bc))+(1/(ac)))


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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

	$We have :$ $\frac{1}{a^{2} + bc} + \frac{1}{b^{2} + ac} + \frac{1}{c^{2} + ab} ⩽ \frac{1}{2} (\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac})$ $\Leftrightarrow \frac{1}{a^{2} + bc} + \frac{1}{b^{2} + ac} + \frac{1}{c^{2} + ab} ⩽ \frac{a + b + c}{2 abc} (1)$ $Applying {Cauchy}^{'} s inequality for two$ $positive numbers we have :$ $a^{2} + bc ⩾ 2 a \sqrt{bc}, b^{2} + ac ⩾ 2 b \sqrt{ac}$ $c^{2} + ab ⩾ 2 \sqrt{ab}, it follows that$ $L . H . S ⩽ \frac{1}{2 a \sqrt{bc}} + \frac{1}{2 b \sqrt{ac}} + \frac{1}{2 c \sqrt{ab}}$ $Hence, we just need prove that :$ $\frac{1}{2 a \sqrt{bc}} + \frac{1}{2 b \sqrt{ac}} + \frac{1}{2 c \sqrt{ab}} ⩽ \frac{a + b + c}{2 abc} (2)$ $\Leftrightarrow \frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} ⩽ \frac{a + b + c}{\sqrt{abc}}$ $\Leftrightarrow \sqrt{ab} + \sqrt{bc} + \sqrt{ca} ⩽ a + b + c$ $\Leftrightarrow 2 (\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) ⩽ 2 (a + b + c)$ $\Leftrightarrow {(\sqrt{a} - \sqrt{b})}^{2} + {(\sqrt{b} - \sqrt{c})}^{2} + {(\sqrt{c} - \sqrt{a})}^{2} ⩾ 0$ $This inequality is always true$ $\forall a, b, c > 0 . Hence, (2) is true infer (1) is$ $true . Thus, the given inequality is$ $proved . The equality ocurrs if and only$ $if a = b = c . (Q . E . D)$
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