Show that for all real numbers $a$ and $b$, $\,\, ab \le (1/2)(a^2+b^2)$

• Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$

• Simple algebra question – proving $a^2+b^2 \geqslant 2ab$ [duplicate]

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Hint: $\forall a,b \in \mathbb R, \; (a-b)^2\ge 0$

Hint: Think about expanding and rearranging the inequality $(a-b)^2\geq 0$

Try using polar coordinates

$$a=r\cos \theta,b=r \sin \theta.$$

The inequality then becomes $$r^2 \cos \theta \sin \theta \leq \frac{1}{2}r^2,$$
or equivalently for nonzero $r$ $$2 \cos \theta \sin \theta (=\sin 2 \theta) \leq 1 .$$

We have a theorem in Classical Algebra that the arithmetic mean of n positive real numbers is greater than or equal to their geometric mean, the equality occurs when the numbers are equal.
In short, $A.M \geq G.M$

For any two real numbers $a, b,$ we know that $a^2,b^2$ are two positive real numbers. Now we apply the above stated theorem here.
$${{a^2+b^2}\over 2} \geq (a^2b^2)^{1\over2}\\ or,\space {{a^2+b^2}\over 2} \geq ab\\or, \space ab\leq{{a^2+b^2}\over 2}$$
This completes the proof.