Problem
ALG-B2-M08-P003 Roots as squares
#3
★★☆☆☆ Level 2 of 5
Prove for \(a,b\ge0\): \(\sqrt{a}+\sqrt{b}\le\sqrt{2(a+b)}\).
Hint. Set \(a=x^2\), \(b=y^2\).
Let \(a=x^2\), \(b=y^2\), \(x,y\ge0\). We need \(x+y\le\sqrt{2(x^2+y^2)}\). Squaring gives \(2xy\le x^2+y^2\), i.e. \((x-y)^2\ge0\).
Teaching goal: recognize the appropriate substitution and check the domain.