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Algebra Level 4

$\sqrt{x + 672} + \sqrt{y + 672} + \sqrt{z + 672}$

Let $x,y$ and $z$ be positive reals whose sum is 2016. If the maximum value of the expression above is $M$ , what is $M^2 ?$

2 solutions

Dev Sharma
Dec 28, 2015

We will use Cauchy Schwarz Inequality,

$(1 + 1 + 1)((x + 672) + (y + 672) + (z + 672)) ≥ (\sqrt{x + 672} + \sqrt{y + 672} + \sqrt{z + 672})^2$

$3(x + y + z + 2016) ≥ (\sqrt{x + 672} + \sqrt{y + 672} + \sqrt{z + 672})^2$

$12096 ≥ (\sqrt{x + 672} + \sqrt{y + 672} + \sqrt{z + 672})^2$

$\sqrt{12096} ≥ (\sqrt{x + 672} + \sqrt{y + 672} + \sqrt{z + 672})$

So maximum value is $M = \sqrt{12096}$ , thus $M^2 = 12096$ .

In your question you are asking for the minimum value. Please edit it.

Kushagra Sahni - 5 years, 5 months ago

Done. Thanks.

Dev Sharma - 5 years, 5 months ago

Rishik Jain
Jan 5, 2016

We know that $A.M \le R.M.S$

$\dfrac{\sqrt{x+672}+ \sqrt{y+672}+ \sqrt{z+672}}{3} \le \sqrt{\dfrac{x+672+y+672+z+672}{3}}$

$\sqrt{x+672}+ \sqrt{y+672}+ \sqrt{z+672} \le 3 \sqrt{\dfrac{4032}{3}}$

$\sqrt{x+672}+ \sqrt{y+672}+ \sqrt{z+672} \le 3 \sqrt{1344}$

Therefore, the maximum value $M= 3 \sqrt{1344}$

$M^2 = 9\times 1344 = \large \boxed{12096}$