Variation on Pythagoras: If $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$

Proof without words:

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Assuming you are looking for $a+b\le c\sqrt 2$, as a hint, consider $(a+b)^2+(a-b)^2$

Let $x=(a,b)$ and $y=(1,1)$, then we have:

$a+b=x\centerdot y\leq||x||\space||y||=\sqrt{a^2+b^2}\sqrt{2}=c\sqrt{2}$.

$(a+b)^2\\ \leq(a+b)^2+(a-b)^2\\ =a^2+2ab+b^2+a^2-2ab+b^2\\ =2a^2+2b^2 =2(a^2+b^2) =2c^2$

Using the square root gives us:

$a+b\leq\sqrt{2}c$

You have $c=\sqrt{a^2+b^2}$, so your inequality is equivalent to $$\frac{a+b}2 \le \sqrt{\frac{a^2+b^2}2}.$$ This is the well known inequality between quadratic and arithmetic mean.

In this case we only need two variables, but it is true for more variables, too.

If $x_1,\dots,x_n\ge0$ are real numbers then $$\frac{x_1+x_2+\dots+x_n}n \le \sqrt{\frac{x_1^2+x_2^2+\dots+x_n^2}n}.$$ The equality holds if and only if $x_1=x_2=\dots=x_n$.

Some links:

• Prove QM-AM inequality
• Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at AoPS
• Inequalities: A Mathematical Olympiad Approach By Radmila Bulajich Manfrino et al., p.36