Can Inequality be Proven for Positive Reals a and b?

[anemone]

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

Spoiler

Squaring the LHS we have

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

Which may be written as

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

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

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

Now,

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

Focussing on the numerator,

\(\displaystyle ab(a+b)^2-2ab(a^2+b^2)=a^3b+2a^2b^2+ab^3-2a^3b-2ab^3\)
\(\displaystyle =ab(2ab-a^2-b^2)\)

Now consider

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

\(\displaystyle a^2-2ab+b^2\ge0\)

\(\displaystyle a^2+b^2\ge2ab\)

hence

\(\displaystyle ab(2ab-a^2-b^2)\)

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

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

and the original inequality follows.

 

[anemone]

Spoiler

Squaring the LHS we have

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

Which may be written as

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

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

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

Now,

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

Focussing on the numerator,

\(\displaystyle ab(a+b)^2-2ab(a^2+b^2)=a^3b+2a^2b^2+ab^3-2a^3b-2ab^3\)
\(\displaystyle =ab(2ab-a^2-b^2)\)

Now consider

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

\(\displaystyle a^2-2ab+b^2\ge0\)

\(\displaystyle a^2+b^2\ge2ab\)

hence

\(\displaystyle ab(2ab-a^2-b^2)\)

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

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

and the original inequality follows.

Very well done, greg1313!