If a-b-c - R- such that a-b-c-3- then the greatest value of -4a-1-4b-1-4c-1 is

The correct option is A 9 Given that a+b+c=3 we know that A.M.≥ G.M. x+y2≥√(xy) let x=4a+1, y=1 therefore, #160; 4a+1+12≥√((4a+1) ...

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Standard IX

Mathematics

Special Products

If a,b,c ∈ ...

Question

If $(a, b, c) \in R^{+}$ such that $a + b + c = 3$ , then the greatest value of $\sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1}$ is

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Solution

The correct option is A $9$
Given that $a + b + c = 3$
we know that $A . M . \geq G . M .$

$⟹ \frac{x + y}{2} \geq \sqrt{x y}$

let $x = 4 a + 1, y = 1$

therefore, $\frac{4 a + 1 + 1}{2} \geq \sqrt{(4 a + 1) (1)}$

$⟹ \frac{4 a + 2}{2} \geq \sqrt{4 a + 1}$

$⟹ 2 a + 1 \geq \sqrt{4 a + 1}$

$⟹ \sqrt{4 a + 1} \leq 2 a + 1$ --------(1)

similarly, $\sqrt{4 b + 1} \leq 2 b + 1$ --------(2)

and $\sqrt{4 c + 1} \leq 2 c + 1$ --------(3)

adding (1), (2) and (3) we get

$\sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1} \leq 2 a + 1 + 2 b + 1 + 2 c + 1$

$⟹ \sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1} \leq 2 (a + b + c) + 3$

$⟹ \sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1} \leq 2 (3) + 3$

$⟹ \sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1} \leq 9$

therefore, the greatest value of $\sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1}$ is $9$

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If (a,b,c)∈R+ such that a+b+c=3, then the greatest value of √4a+1+√4b+1+√4c+1 is

If $(a, b, c) \in R^{+}$ such that $a + b + c = 3$ , then the greatest value of $\sqrt{4 a + 1} + \sqrt{4 b + 1} + \sqrt{4 c + 1}$ is