Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
Already signed up?
New to KöMaL?
I want the old design back!!! :-)

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 April 10, 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

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

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

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


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

Problems in Mathematics of KöMaL, March 2017