One classical trick is the following:

Since

[tex] (x -y)^2 \ge 0, [/tex] we have

[tex] x^2 – 2 x y + y^2 \ge 0 \implies x^2 + 2 x y + y^2 \ge 4xy [/tex]

Taking root we obtain [tex] (x+y)/2 \ge \sqrt{xy} [/tex]

A variation of this trick can be used to show that

[tex]| a \sin x + b \cos x | \le \sqrt{a^2 +b^2}. [/tex]

We start with [tex](a\cos x – b \sin x)^2 \ge 0. [/tex] Expanding the expression we get

[tex] a^2 – a^2 \sin^2 x – 2 a b \sin x \cos x + b^2 – b^2 \cos^2 x \ge 0. [/tex]

A simple rearrangement completes the proof.

i realize this is an old post… but looking at this from a justify-each-step point of view… I came to the conclusion that the following condition must be true: x,y>1… but at the same time, I wondered about cases. What you’ve said is true for 1) x,y>1, but what about 2) x,y1>y>0 (that’s assuming x,y>0). Instinct tells me that case 3) depends on if x-y >0.

Then I checked a few of my assumptions… only to find out that they were all wrong.

My confusion all stemmed from the fact that I accepted that IF a<b, THEN a^2<b^2… BUT I rejected IF a^2<b^2, THEN a<b (necessarily). I didn’t accept the second conditional because if x<1, then x^2<x<1.

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So… my point is… thanks for the simple, but elegant proofs that made me run a few mental and arithmetic circles.