Inequality -- 3

Algebra Level 2

Positive reals $a$ and $b$ are such that $a+b=5$ . What is the maximum value of $\sqrt {a+1}+\sqrt {b+3}$ ?

3\sqrt 2

3\sqrt 3

5 4

2 solutions

Chew-Seong Cheong
Sep 7, 2019

By Cauchy-Schwarz inequality ,

$\begin{aligned} (\sqrt{a+1}+\sqrt{b+3})^2 & \le 2({\color{#3D99F6} a} +1+{\color{#3D99F6} b} +3) =2({\color{#3D99F6} 5}+4 ) \\ \implies \sqrt{a+1}+\sqrt{b+3} & \le \boxed{3\sqrt 2}\end{aligned}$

Equality occurs when $a+1 = b+3$ or $a=\frac 72$ and $b=\frac 32$ .

To complete this proof, you need to show that the upper bound of $3\sqrt{2}$ can be achieved, with $a=\tfrac72$ , $b=\tfrac32$ .

Mark Hennings - 1 year, 9 months ago

Thanks. I will add it in.

Chew-Seong Cheong - 1 year, 9 months ago

Kevin Xu
Sep 6, 2019

Yes, it is true. The information with regard to $a+b = 5$ is useless. $\\$ By AM-GM $\\$ $\sqrt {a+1}+\sqrt {b+3} \leq 2\sqrt {\frac 92}$ $\\$ $\sqrt {a+1}+\sqrt {b+3} \leq 3 \sqrt 2$ $\\$

Inequality -- 3

2 solutions

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