Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 5 down vote accepted

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 $$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$

share|cite|improve this answer
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 $(c-1)^2(a+b) \ge 0$.

share|cite|improve this answer
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 $p=a+b+c$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.