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Problem A. 571. (October 2012)

A. 571. The side lengths of a triangle are a, b, and c. Prove that

\sqrt{\frac{(2a+c)(2b+c)}{(3a+b)(3b+a)}} +
\sqrt{\frac{(2b+a)(2c+a)}{(3b+c)(3c+b)}} +
\sqrt{\frac{(2c+b)(2a+b)}{(3c+a)(3a+c)}} < \frac52.

Proposed by: Daniel Campos, San Jose, Costa Rica

(5 pont)

Deadline expired on November 12, 2012.

Solution. Notice that we have equality in the degenerate case when one of a,b,c is zero and the other two are equal.

We show that

\sqrt{\frac{(2a+c)(2b+c)}{(3a+b)(3b+a)}} < \frac{a+b+3c}{2(a+b+c)}.

(This estimate preserves equality in the case mentioned.)


\left(\frac{a+b+3c}{2(a+b+c)}\right)^2 - \frac{(2a+c)(2b+c)}{(3a+b)(3b+a)} =

= \frac{(a+b-c)\Big(8(a+b)^2c+20(a+b)c^2+4c^3+(3a+3b+5c)(a-b)^2\Big)}{4(a+b+c)^2(3a+b)(3b+a)}
> 0.

By the cyclic permutations of the variables we get \sqrt{\frac{(2b+a)(2c+a)}{(3b+c)(3c+b)}}<\frac{3a+b+c}{2(a+b+c)} and \sqrt{\frac{(2c+b)(2a+b)}{(3c+a)(3a+c)}}<\frac{a+3b+c}{2(a+b+c)}. The sum of these and (1) is the statement of the problem.


11 students sent a solution.
5 points:Bodnár Levente, Cyril Letrouit, Herczeg József, Janzer Olivér, Kúsz Ágnes, Maga Balázs, Nagy Róbert, Omer Cerrahoglu, Williams Kada.
0 point:2 students.

Problems in Mathematics of KöMaL, October 2012