Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$

Question

Question:

Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.

So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to definitively say that the statement is proved.

This is my "work" so far:

$a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$

$\frac {2a(b + 2a) + 2b(a + 2b) - 1(b + 2a)(a + 2b)}{2(a + 2b)(b+2a)} ≥ 0$

$\frac {4ab + 4a^2 + 4b^2 - 2b^2 - 5ab - 2a^2}{4b^2 + 10ab + 4a^2} ≥ 0$

$\frac {2a^2 - ab + 2b^2}{4b^2 + 10ab + 4a^2} ≥ 0$

Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Martin R, Aug 25 '22 at 17:38
You need that $0 \leq (a-b)^2 \implies (a^2 + b^2) \geq 2ab.$ — user2661923, Aug 25 '22 at 17:58
Now you have shown your work... the denominator must be >0, so you just need to show that the numerator >0. — Blitzer, Aug 25 '22 at 18:07
@jjagmath Thank you for noticing & letting me know. I misread the inequality there initially thinking it was the same one. To help avoid any potential confusion, I've deleted my comment with the link. — John Omielan, Aug 25 '22 at 19:45

score 2 · Accepted Answer · answered Aug 25 '22 at 18:13

2

You almost have it, just use

$$2a^2-ab+2b^2 = 2(a-b)^2+3ab > 0$$

(and the denominator is positive as well).

answered Aug 25 '22 at 18:13

jjagmath

18,214

Yes, this is probably the answer that at my stage of learning is most appropriate, thank you. – Jamie Aug 25 '22 at 18:36
@Jamie You could accept my answer if you found it helpful. – jjagmath Aug 25 '22 at 19:23

score 0 · Answer 2 · answered Aug 25 '22 at 18:04

One of $a$ or $b$ is non zero, or else the question is not well defined.

If $a=b$, then we get left hand side of inequality to be $2/3$, which is $\geq 1/2$ as desired.

So assume $a>b$. Then $$ \frac{a}{a+2b} + \frac{b}{b + 2a} \\ = \frac{a(b+2a)+b(a+2b)}{(a+2b)(b+2a)} \\ = \frac{2ab+2a^2+2b^2}{3ab + (2ab+2a^2 + 2b^2)} \\ = \frac{x}{3ab + x} \mbox{ where } x = 2ab+2a^2 + 2b^2 > 0 \\ = \frac{1}{\frac{3ab}{x} + 1} $$ So we need to show that $$\frac{3ab}{x} \leq 1 \iff 3ab \leq x \iff ab \leq 2a^2+2b^2 $$ But the furthest right inequality is obvious since $$ ab < a^2 < 2a^2 < 2a^2 + 2b^2 $$

score 0 · Answer 3 · answered Aug 25 '22 at 18:20

first let's do some cleaning: $$\frac{a}{a+2b}+\frac{b}{b+2a} = \frac{2a^2 + 2ab + 2b^2}{2a^2 + 4ab + 2a^2} = \frac{2(a+b)^2 - 2ab}{2(a+b)^2} = 1-\frac{ab}{(a+b)^2}$$ and we have : $$A = \frac{a}{a+2b}+\frac{b}{b+2a} = \frac{1}{1+\frac{2b}{a}} + \frac{1}{1+\frac{2a}{b}}$$ let $x = \frac{a}{b} $ then we have : $$f(x) = \frac{1}{1+\frac{2}{x}} + \frac{1}{1+2x} = \frac{x}{x+2} + \frac{1}{1+2x}$$ therefore: $$f^\prime(x) = \frac{2}{(x+2)^2} - \frac{2}{(1+2x)^2} = 0 \Rightarrow x^2 + 4x + 4 = 4x^2 + 4x + 1\Rightarrow 3x^2-3 = 0 \Rightarrow x = \pm 1 \Rightarrow f(x) = \frac{2}{3}, -2$$ but since both a and b are positive x = -2 is not an option and therefore we have that with x>0 f(x) is always larger than 2/3 which is a subset of the desired result.$\blacksquare$

Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$

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