MHB Can Inequality be Proven for Positive Reals a and b?

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The discussion focuses on proving the inequality $$\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}$$ for all positive real numbers a and b. Participants express initial confidence in the problem's solvability, but one contributor notes the challenge posed by the original poster's reputation. The conversation highlights the complexity of the inequality despite its seemingly straightforward appearance. The proof is acknowledged as well-executed by a participant named greg1313. The thread emphasizes the mathematical exploration of inequalities involving positive reals.
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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 $b$.
 
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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
 
anemone said:
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 $b$.

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.
 
greg1313 said:
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!:cool:
 
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