Can Inequality be Proven for Positive Reals a and b?

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Discussion Overview

The discussion centers around 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 are exploring the validity and potential proof of this mathematical statement.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Post 1 presents the inequality as a statement to be proven for positive reals $a$ and $b$.
  • Post 2 expresses a personal reaction to the problem, indicating initial confidence in its simplicity but also acknowledging the reputation of the poster.
  • Post 3 reiterates the inequality, suggesting a focus on its proof.
  • Post 4 offers praise to another participant, indicating a positive reception of their contributions, though it does not clarify what specific points were well done.

Areas of Agreement / Disagreement

The discussion does not show clear agreement or disagreement on the validity of the inequality itself, as the focus appears to be on the challenge of proving it rather than on differing viewpoints regarding its truth.

Contextual Notes

No specific limitations or assumptions are discussed, but the nature of the inequality suggests that its proof may depend on various mathematical techniques or properties of inequalities.

Who May Find This Useful

Participants interested in mathematical inequalities, proof techniques, or those studying properties of positive real numbers may find this discussion relevant.

anemone
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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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