Magyar Information Contest Journal Articles

# Problem A. 694. (March 2017)

A. 694. Prove that the inequality

$\displaystyle \frac1{\sqrt{2x}} + \frac1{\sqrt{2y}} + \frac2{\sqrt{x+y}} + 2 \ge \frac4{\sqrt{x+2}} + \frac4{\sqrt{y+2}}$

holds for all pairs $\displaystyle (x,y)$ of positive real numbers.

(5 pont)

Deadline expired on 10 April 2017.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. A feladat az A. 493. feladat speciális esete az $\displaystyle n=3$, $\displaystyle p_1=x$, $\displaystyle p_2=y$, $\displaystyle p_3=2$, $\displaystyle a_1=a_2=1$, $\displaystyle a_3=-2$, $\displaystyle c=\tfrac12$ választással.

Bővebben: bármely $\displaystyle p>0$ esetén, az $\displaystyle u=pt$ helyettesítéssel

$\displaystyle \int_{t=0}^\infty \frac{e^{-pt}}{\sqrt{t}} \mathrm{d}t = \int_{u=0}^\infty \frac{e^{-u}}{\sqrt{u/p}} \frac{\mathrm{d}t}{p} = \frac{1}{\sqrt{p}}\int_{u=0}^\infty \frac{e^{-u}}{\sqrt{u}} \mathrm{d}u = \frac{\Gamma(\tfrac12)}{\sqrt{p}},$

ezt felhasználva

$\displaystyle B.O.-J.O. = \frac1{\sqrt{2x}} +\frac2{\sqrt{x+y}} +\frac1{\sqrt{2y}} -\frac4{\sqrt{x+2}} -\frac4{\sqrt{y+2}} +\frac{4}{\sqrt4} =$

$\displaystyle = \frac{1}{\Gamma(\tfrac12)} \int_0^\infty \Big( e^{-2xt}+2e^{-(x+y)t}+e^{-2yt}-4e^{-(x+2)t}-4e^{-(y+2)t}+4e^{-4t} \Big)\frac{\mathrm{d}t}{\sqrt{t}} =$

$\displaystyle = \frac{1}{\Gamma(\tfrac12)} \int_0^\infty \Big( e^{-xt}+e^{-yt}-2e^{-2t}\Big)^2 \frac{\mathrm{d}t}{\sqrt{t}} \ge 0.$

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 5 students sent a solution. 5 points: Williams Kada. 0 point: 4 students.

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