Given that a and b are positive real number such that b<4a+1,show that ((2a+b)/(4a+1))<(√(4a^2 +b)) .


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Question Number 6257 by 314159 last updated on 20/Jun/16

$G i v e n t h a t a a n d b a r e p o s i t i v e r e a l n u m b e r$ $s u c h t h a t b < 4 a + 1, s h o w t h a t \frac{2 a + b}{4 a + 1} < \sqrt{4 a^{2} + b} .$
	Answered by Yozzii last updated on 20/Jun/16

	$\frac{2 a + b}{4 a + 1} < \sqrt{4 a^{2} + b} a, b > 0$ $\Rightarrow \frac{{(2 a + b)}^{2}}{{(4 a + 1)}^{2}} < 4 a^{2} + b$ $- - - - - - - - - - - - - - - - - - - - - - - - - - -$ $L e t ϕ = \frac{{(2 a + b)}^{2}}{{(4 a + 1)}^{2}} - 4 a^{2} - b .$ $ϕ = \frac{4 a^{2} + 4 a b + b^{2} - (16 a^{2} + 8 a + 1) (4 a^{2} + b)}{{(4 a + 1)}^{2}}$ $ϕ = \frac{4 a^{2} + 4 a b + b^{2} - 64 a^{4} - 16 a^{2} b - 32 a^{3} - 8 a b - 4 a^{2} - b}{{(4 a + 1)}^{2}}$ $ϕ = \frac{- 64 a^{4} - 16 a^{2} b - 32 a^{3} - 4 a b + b (b - 1)}{{(4 a + 1)}^{2}}$ $S i n c e b < 4 a + 1 \Rightarrow b - 1 < 4 a \Rightarrow b (b - 1) < 4 a b$ $\Rightarrow ϕ < \frac{- 64 a^{4} - 16 a^{2} b - 32 a^{3} - 4 a b + 4 a b}{{(4 a + 1)}^{2}}$ $ϕ < \frac{- 16 a^{2} (4 a^{2} + b + 2 a)}{{(4 a + 1)}^{2}}$ $S i n c e a, b > 0 \Rightarrow 4 a^{2} + 2 a + b > 0, - 16 a^{2} < 0$ $a n d {(4 a + 1)}^{2} > 0 . T h u s, ϕ < 0 .$ $\Rightarrow 0 < \frac{{(2 a + b)}^{2}}{{(4 a + 1)}^{2}} < 4 a^{2} + b$ $∴ \sqrt{{(\frac{2 a + b}{4 a + 1})}^{2}} < \sqrt{4 a^{2} + b}$ $\Rightarrow∣ \frac{2 a + b}{4 a + 1} ∣< \sqrt{4 a^{2} + b}$ $a, b > 0 \Rightarrow 2 a + b > 0 a n d 4 a + 1 > 0 .$ $∴ \frac{2 a + b}{4 a + 1} < \sqrt{4 a^{2} + b} .$
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