Prove that -a-cb-d-ab-cd a-b-c and d -0

(a+b) 2 ≥√( ab ) (c+d) 2 ≥√( cd ) Adding both we get, a+b+c+d 2 ≥( √( ab ) +√( cd ) #160; ) (1) (a+c)+(b+d) 2 ≥√( (a+c)(b+d) ) ...

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Standard XII

Mathematics

Derivative from First Principle

Prove that ...

Question

Prove that $\sqrt{(a + c) (b + d)} \geq \sqrt{a b} + \sqrt{c d}$ (a.b.c and d >0)

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Solution

$\frac{(a + b)}{2} \geq \sqrt{a b} \frac{(c + d)}{2} \geq \sqrt{c d}$
Adding both we get,
$\frac{a + b + c + d}{2} \geq (\sqrt{a b} + \sqrt{c d}) (1) \frac{(a + c) + (b + d)}{2} \geq \sqrt{(a + c) (b + d)} (2)$
Dividing $(1)$ by $(2)$
$1 \geq \frac{\sqrt{a b} + \sqrt{c d}}{\sqrt{(a + c) (b + d)}} \sqrt{(a + c) (b + d)} \geq \sqrt{a b} + \sqrt{c d}$

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Prove that √(a+c)(b+d)≥√ab+√cd (a.b.c and d >0)

(a+b)2≥√ab(c+d)2≥√cdAdding both we get, a+b+c+d2≥(√ab+√cd)(1)(a+c)+(b+d)2≥√(a+c)(b+d)(2)Dividing (1) by (2)1≥√ab+√cd√(a+c)(b+d)√(a+c)(b+d)≥√ab+√cd

Prove that $\sqrt{(a + c) (b + d)} \geq \sqrt{a b} + \sqrt{c d}$ (a.b.c and d >0)