A problem

Algebra Level 3

If $a+b+c =6,$ where $a, b$ and $c$ are real numbers greater than or equal to $- \frac{1}{4},$ what is the range of the following expression:

$\sqrt{4a +1} + \sqrt{4b +1} + \sqrt{4c+1}?$

\geq 12

\leq 9

10

1 solution

Otto Bretscher
Nov 20, 2015

By Jensen's inequality, applied to the concave function $\sqrt{x}$ , the given expression is $\leq 3\sqrt{\frac{(4a+1)+(4b+1)+(4c+1)}{3}}=3\sqrt{\frac{4*6+3}{3}}=9$ . Equality holds when $a=b=c=2$ .

(Jensen's inequality is essentially the definition of concavity.)

A problem

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