Can Inequality be Proven for Positive Reals a and b?

[anemone]

Prove that $$\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}$$ for all positive reals $a$ and $b$.

 

[topsquark]

You know, I just looked at this and said to myself "Oh! This doesn't look so hard." (Thinking) Then I noted who posted it. (Doh)

-Dan

 

[Greg]

Prove that $$\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}$$ for all positive reals $a$ and $b$.

Spoiler

Squaring the LHS we have

$$\dfrac{a^2+b^2}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}$$

Which may be written as

$$1-\dfrac{2ab}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}$$

From th AM-GM inequality, $2\dfrac{\sqrt{ab}}{a+b}$ has an upper bound of $1$, so we have

$$1-\dfrac{2ab}{(a+b)^2}+1+\dfrac{ab}{a^2+b^2}$$

Now,

$$\dfrac{ab}{a^2+b^2}-\dfrac{2ab}{(a+b)^2}=\dfrac{ab(a+b)^2-2ab(a^2+b^2)}{(a^2+b^2)(a+b)^2}$$

Focussing on the numerator,

$$ab(a+b)^2-2ab(a^2+b^2)=a^3b+2a^2b^2+ab^3-2a^3b-2ab^3$$
$$=ab(2ab-a^2-b^2)$$

Now consider

$$(a-b)^2\ge0$$

$$a^2-2ab+b^2\ge0$$

$$a^2+b^2\ge2ab$$

hence

$$ab(2ab-a^2-b^2)$$

has an upper bound of $0$ so we may state

$$\dfrac{a^2+b^2}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}\le2$$

and the original inequality follows.

 

[anemone]

Spoiler

Squaring the LHS we have

$$\dfrac{a^2+b^2}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}$$

Which may be written as

$$1-\dfrac{2ab}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}$$

From th AM-GM inequality, $2\dfrac{\sqrt{ab}}{a+b}$ has an upper bound of $1$, so we have

$$1-\dfrac{2ab}{(a+b)^2}+1+\dfrac{ab}{a^2+b^2}$$

Now,

$$\dfrac{ab}{a^2+b^2}-\dfrac{2ab}{(a+b)^2}=\dfrac{ab(a+b)^2-2ab(a^2+b^2)}{(a^2+b^2)(a+b)^2}$$

Focussing on the numerator,

$$ab(a+b)^2-2ab(a^2+b^2)=a^3b+2a^2b^2+ab^3-2a^3b-2ab^3$$
$$=ab(2ab-a^2-b^2)$$

Now consider

$$(a-b)^2\ge0$$

$$a^2-2ab+b^2\ge0$$

$$a^2+b^2\ge2ab$$

hence

$$ab(2ab-a^2-b^2)$$

has an upper bound of $0$ so we may state

$$\dfrac{a^2+b^2}{(a+b)^2}+2\dfrac{\sqrt{ab}}{a+b}+\dfrac{ab}{a^2+b^2}\le2$$

and the original inequality follows.

Very well done, greg1313!