# If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.

The whole problem is in the title. If you wanna hear what I've tried, well, I've tried multiplying both sides by 3 and then using the homogenic mean. $${3 \over a+b+1} \le \sqrt[3]{{1\over ab}} = \sqrt[3]{c}$$ By adding the inequalities I get $${3 \over a+b+1} + {3 \over b+c+1} + {3 \over c+a+1} \le \sqrt[3]a + \sqrt[3]b + \sqrt[3]c$$ And then if I proof that that is less or equal to 3, then I've solved the problem. But the thing is, it's not less or equal to 3 (obviously, because you can think of a situation like $a=354$, $b={1\over 354}$ and $c=1$. Then the sum is a lot bigger than 3).

So everything that I try doesn't work. I'd like to get some ideas. Thanks.

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Since others have already posted their successful solutions, I need not have it repeated. I just want to point out that your homogenic mean approach will not get you anywhere. This is because if you plug in your suggested samples, you will find $\dfrac {3} {356} less-than-or-equal-to 1$. This means the comparison is NOT fair. That is, either LHS is too small or hte RHS is too large. Hence, you need to find something smaller for your RHS. – Mick Dec 14 '13 at 17:43

let $$a=x^3,b=y^3,c=z^3\Longrightarrow xyz=1$$ since $$y^3+z^3\ge y^2z+yz^2$$ so $$\dfrac{1}{1+b+c}=\dfrac{xyz}{xyz+y^3+z^3}\le\dfrac{xyz}{xyz+y^2z+yz^2}=\dfrac{x}{x+y+z}$$ so $$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$

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Where does the inequality $y^3+z^3 \geq y^2 z + y z^2$ come from? – Aaron Dec 28 '13 at 21:01
@Aaron: $y^3+z^3-y^2z-yz^2=(y^2-z^2)(y-z)=(y+z)(y-z)^2 \geq 0$ – sdcvvc Dec 28 '13 at 21:33
@Aaron It is also enough to assume (WLOG) that $y\ge z$ and then to use the rearrangement inequality. – user26486 Jun 27 '14 at 22:19

For my earlier comment: By expanding everything I mean, you can clear the denominator, write down everything in terms of symmetric polynomials, and try to use AM-GM to compare them.

On the other hand, there is also a one liner, similar to math110's solution:

$$\frac{1}{a+b+1} \leq \frac{2c+ab}{2(a+b+c)+ab+bc+ca}$$

After clearing the denominator, this is equivalent to $(c-1)^2(a+b) \ge 0$.

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It's very nice +1. – math110 Dec 14 '13 at 10:28
Do you mean that I should multiply both sides by $(a+b+1)(b+c+1)(c+a+1)$? I then get that $$a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2- 2a - 2b - 2c \ge 0.$$ But I can't see how I could continue. – user26486 Dec 14 '13 at 15:07
@mathh, if you know Muirhead's inequality, at the final step you just homogenize your inequality (i.e. multiply $(abc)^{2/3}$ to $a+b+c$ part) and use Muirhead. If not, then try to find a clever AM-GM: for example: $a^2b+a^2c \ge 2a^{3/2}$ by AM-GM, then you can use Power mean or other approaches to prove $a^{3/2} + b^{3/2} + c^{3/2} \ge a+b+c$. – user27126 Dec 14 '13 at 18:51
The Power mean inequality says that: $$\left(\frac{a^{3/2}+b^{3/2}+c^{3/2}}{3}\right)^{3/2}\ge\frac{a+b+c}{3}$$right? But how does this help us prove that: $$a^{3/2}+b^{3/2}+c^{3/2}\ge a+b+c?$$ Well, we could divide everything by 3: $$\frac{a^{3/2}+b^{3/2}+c^{3/2}}{3}\ge\frac{a+b+c}{3}$$ And raise everything to the power of $\frac{3}{2}$:$$\left(\frac{a^{3/2}+b^{3/2}+c^{3/2}}{3}\right)^{3/2}\ge\left( \frac{a+b+c}{3} \right)^{3/2}$$ So we have left to prove that: $$\left(\frac{a+b+c}{3}\right)^{3/2}\ge\frac{a+b+c}{3}$$ – user26486 Dec 23 '13 at 22:57
We divide everything by $\frac{a+b+c}{3}$: $$\sqrt{\frac{a+b+c}{3}}\ge1$$ Which is equivalent to: $$a+b+c\ge3$$ But we don't know if that's true. It's not necessarily correct. – user26486 Dec 23 '13 at 22:57

I have other nice Cauchy-Schwarz inequality solve it.

since $$\dfrac{1}{1+a+b}=1-\dfrac{a+b}{1+a+b}$$ so the original inequality can be written $$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge2$$ use Cauchy-Schwarz inequaliy and the AM-GM inequality,we have $$\sum_{cyc}\dfrac{a+b}{a+b+1}\ge\dfrac{(\sum\sqrt{a+b})^2}{\sum(a+b+1)}=\dfrac{2p+2\sum\sqrt{(a+b)(a+c)}}{2p+3}\ge\dfrac{2p+2\sum(a+\sqrt{bc})}{2p+3}=\dfrac{4p+2\sum\sqrt{bc}}{2p+3}\ge 2$$ because use AM-GM inequality $$\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\ge 3\sqrt[3]{abc}=3$$ where $p=a+b+c$

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