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Possible solution

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Question

Is my answer to the problem valid? would there be a more direct way, that is, without building the additional triangle?

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Pierre
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    this is a particular case of Cauchy-Schwarz (inequality and equality case): $ |\binom ab\cdot \binom 11| \le |\binom ab| |\binom 11|$ – Calvin Khor Apr 25 '22 at 08:57
  • sorry but i don't understand the connection – Pierre Apr 25 '22 at 09:47
  • Way with additional triangle is direct and it does not requires any calculation, because hypotenuse of right triangle is always greater than any other side, so for picture 1: $c\sqrt{2}>a+b$, for picture 2 $c\sqrt{2}=a+b$. Other side is $a+b$ because figure formed by additional line and square border is rectangle. – Ivan Kaznacheyeu Apr 25 '22 at 12:44
  • How would you do without the additional triangle, to prove that the images are the solution to the problem? – Pierre Apr 25 '22 at 14:10

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Your answer is correct, but you need to be careful to argue that the angle between the two hypothenuses of length $c$ is a right angle (trivial, but if I was grading this, I would consider it mandatory for a full solution). The idea of constructing the second triangle is quite clever, as is typical for clean solutions for Olympiad problems.
However, if you do know a bit about basic inequalities, a much more down to earth solution is possible. Algebraically, the question can be rephrased as follows. Given $a,b>0$, show that $\sqrt{2}c\geq a+b$, where $c=\sqrt{a^2+b^2}$. Squaring everything, you need to show that $(a+b)^2\leq 2(a^2+b^2)$ or equivalently that $a^2+b^2\geq 2ab$ or $(a-b)^2\geq 0$, which is always true and equality holds if and only if $a=b$.

The proposed solution is just the geometric interpretation of this inequality as you demonstrate.

As Calvin Khor said, this can also be seen as an application of the Cauchy-Schwarz inequality, but that is a bit overkill in my opinion.

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