Track / Stage 4 / 263 of 340 #523 of 1964

Problem 523

AMC 12 late, AIME early

Algebra Difficulty 4.9 Multiple choice

6. Given $x, y, t \in \mathbf{R}_{+}$ , let
$M=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}, N=\frac{4(x+y)}{(x+y)^{2}+4 t} \text {. }$

Then among the following 4 propositions:
(1) If $x y=3 t$ , then $M=N$ ;
(2) If $x y>3 t$ , then $M \leqslant N$ ;
(3) If $x y<3 t$ , then $M \geqslant N$ ;
(4) If $x y=3 t, M=N$ does not necessarily always hold, the number of correct propositions is:

Pick one

Official solution

6. A.
$\begin{array}{l} M-N=\frac{x}{x^{2}+t}+\frac{y}{y^{2}+t}-\frac{4(x+y)}{(x+y)^{2}+4 t} \\ =\frac{(x+y)(x y+t)}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}-\frac{4(x+y)}{(x+y)^{2}+4 t} . \end{array}$

Let $P=\frac{x y+t}{x^{2} y^{2}+\left(x^{2}+y^{2}\right) t+t^{2}}$ .
Substituting $x y=3 t$ into the above equation, we get
$P=\frac{4}{x^{2}+y^{2}+10 t}=\frac{4}{(x+y)^{2}+4 t} \text {. }$

Therefore, when $x y=3 t$ , (1) is correct, and (1) is not incorrect.
Let $x=2 \sqrt{t}, y=3 \sqrt{t}$ , then $x y>3 t$ , and
$M=\frac{7}{10 \sqrt{t}}>\frac{20}{29 \sqrt{t}}=N \text {. }$

Let $x=2 \sqrt{t}, y=\sqrt{t}$ , then $x y<3 t$ , and
$M=\frac{9}{10 \sqrt{t}}, N=\frac{12}{13 \sqrt{t}} \text {. }$

Thus, $M<N$ . Therefore, (2), (3), and (4) are all incorrect.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.