Let $a,b,c>0$ and $\frac1{a+1}+\frac1{b+1}+\frac1{c+1}=1.\qquad $ Prove $$\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}+\frac{b}{\sqrt{ca}}\cdot\frac1{b+1}+\frac{c}{\sqrt{ab}}\cdot\frac1{c+1}\leqslant\sqrt2.$$

The full solution (due to @Hanno) see at the very end in $\underline{\textbf{Conclusion}}\,\,$

My attemps:

  1. AM>HM gives $a+b+c\geqslant 6.$ Put $a=a_1t,b=b_1t,c=c_1t, \,t\geqslant 1, \, a_1+b_1+c_1=6$

As $\frac 1{x+1}$ is decreasing function we have $\sum\limits_{cyc}\frac{a}{\sqrt{bc}}\cdot\frac1{a+1}\leqslant \sum\limits_{cyc}\frac{a_1}{\sqrt{b_1c_1}}\cdot\frac1{a_1+1}.$ So is it possible to prove the same inequality with new restriction $a+b+c=6?$

  1. $a=\tan^2\alpha, b=\tan^2\beta, c=\tan^2\gamma$ gives

$$\sum\limits_{cyc}\sin^2\alpha\cot\beta\cot\gamma\leqslant \sqrt2$$ if $0<\alpha,\beta,\gamma<\frac\pi2$ and $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.$

  1. After hint of Michael Rozenberg. Rewrite the substitution as linear system with determinant $d=\begin{vmatrix}-a&1&1\\ 1&-b&1\\ 1&1&-c\\ \end{vmatrix}$

If $d\ne 0$ then $x=y=z=0$ and so the hint doesn't work

The equality $d=0$ gives editional restriction, but I am not sure that this case deserves attention.

In fact, I have an ugly solution with brute force, but I want something beautiful.

  1. Continue. Agree with $a=\frac{y+z}{x}$, $b$ and $c$ cyclically and $x+y+z=1.$ I'll try not to break the symmetry as in my ugly solution.

Substution gives $$\sum\limits_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+z)(x+y)}}\leqslant\sqrt2$$ squaring gives $$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+2\frac{(z+x)\sqrt{zx}}{\sqrt{(y+x)(y+z)}}\cdot\frac{(x+y)\sqrt{xy}}{\sqrt{(z+y)(z+x)}}\right)\leqslant2$$ or $$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+ 2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2$$ I do not see here where to use AM-GM, or did I miss the moment?

Thanks, @Michael Rozenberg, very cool. I did not miss, I did not get to the point, although I was already near. Why I stubbornly did not want to multiply by the common denominator? Although further my restriction $x+y+z=1$ would not help but would only hinders.

  1. Yesterday I did not check the last inequality but about this later.

We conside the case $d=0.$ In order for the linear system to have a unique solution, it is necessary to add a new constraint. I put $x+y+z=1$ and violated homogenity. To save it let $x+y+z=I$ where $I$ denotes the sum only. Then from the system we have $x=\frac I{a+1}$ and $a=\frac Ix-1=\frac{y+z}{x}$ i.e. we simply forget that $I=1$ and save the homogenity.

we amend the previous one: $$\sum\limits_{cyc}\left(\frac{(y+z)^2yz}{(x+z)(x+y)}+ 2\frac{x\sqrt{yz(x+y)(x+z)}}{z+y}\right)\leqslant2I^2$$

After AM-GM we have

$$\sum\limits_{cyc}\left({yz(y+z)^3}+ x(x+y)(x+z)(xy+xz+2yz)\right)\leqslant2I^2\prod\limits_{cyc}(x+y)$$

It remains to check the last inequality.

For this its sufficent to calculete the all summands in both sides. I do not want to apply brutal forse.

Avidently both sides have can have members only of type $x^\alpha y^\beta z^\gamma$ with $\alpha+\beta+\gamma=5$ and by symmetry we may assume $\alpha\leqslant\beta\leqslant\gamma.$ My quation is: Is there another way?

Start with memmber $yz^4$. It is contained in the LHS but not contained in the right. So the last inequality must contain negative term and the proof must be corrected if it is possibble yet. Or I'm wrong.

$\underline{\textbf{Conclusion}}\,\,$ The inequality turned out to be interesting and the proof gave @Hanno. Allow yourself to make some minor corrections.

The original restriction was $$\sum_{\text{cyc}}{1\over a+1} =1 \qquad (1)$$ Now $\sum_{\text{cyc}}{a\over a+1} =\sum_{\text{cyc}}{a+1-1\over a+1} =3-1=2.$ So $$\sum_{\text{cyc}}{a\over a+1} =2 \qquad (2)$$ Apply the Cauchy–Bunyakovsky–Schwarz inequality (CBS):

$$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1} \:= \: \sum_{\text{cyc}}\sqrt{{a\over a+1}}\cdot\sqrt{\frac a{(a+1)bc}}\:\leqslant\: {\sqrt{\sum_{\text{cyc}}{a\over a+1}}}\,\cdot\, \sqrt{\sum_{\text{cyc}}{a\over (a+1)bc}}$$ Taking into account (2) it remains to proof inequality $\sum\limits_{\text{cyc}}{a\over (a+1)bc}\:{\leqslant}\:1$ which is equivalent to

$$\sum_{\text{cyc}}{a^2\over a+1}\:\leqslant\: abc $$ We have $\sum\limits_{\text{cyc}}{a^2\over a+1}=\sum\limits_{\text{cyc}}{a^2-1+1\over a+1}=\sum\limits_{\text{cyc}}(a-1)+\sum\limits_{\text{cyc}}{1\over a+1}=a+b+c-2<a+b+c+2=abc. $

The last equality $a+b+c+2=abc$ follows from (1) by bringing to a common denominator. DONE!

The inequality is exact. Take small $a$ and $b=c=1+\frac2a.$ Then (1) is true and the LHS of required inequality evaluates to ${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}$ which tends to $\sqrt2$ if $a\to0$.

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    $\begingroup$ I get highly suspicious that Cauchy Schwarz inequality might greatly help here $\endgroup$ – Darkrai Mar 19 '18 at 2:02
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    $\begingroup$ Did you try Cauchy-Schwarz inequality with appropriate vectors and appropriate inner product? $\endgroup$ – Tal-Botvinnik Mar 19 '18 at 2:04
  • $\begingroup$ Surely I looked in this direction but did not have any success $\endgroup$ – Minz Mar 19 '18 at 4:41
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    $\begingroup$ Can you offer a reference for this inequality? Where did you come across it? $\endgroup$ – Hanno Mar 24 '18 at 19:45
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    $\begingroup$ @Hanno thanks will consider $\endgroup$ – Minz Mar 26 '18 at 14:31

The presented proof relies alone on the Cauchy–Bunyakovsky–Schwarz inequality (CBS).

We exploit equivalent formulations of the given constraint \begin{align*} & \sum_{\text{cyc}}{1\over a+1} =1 \tag{1} \\[1ex] \iff\quad & \sum_{\text{cyc}}{a\over a+1}=2 \tag{2} \\[1ex] \iff\quad & abc = a+b+c+2 \tag{3} \end{align*} The latter one stems from clearing denominators and follow-up purge.

Upon application of (CBS) and using $(2)$ the upper limit of $\sqrt2$ materialises: $$\sum_{\text{cyc}}{a\over\sqrt{bc}}\cdot\frac1{a+1} \:= \: \sum_{\text{cyc}}\sqrt{{a\over a+1}\cdot\frac a{(a+1)bc}}\:\leqslant\: \underbrace{\sqrt{\sum_{\text{cyc}}{a\over a+1}}}_{\sqrt2}\,\cdot\, \left(\sum_{\text{cyc}}{a\over (a+1)bc}\right)^{1/2}$$ Note the close similarity between the initial expression and the last factor. It remains to show that it is bounded above by $1$ $$\sum_{\text{cyc}}{a\over (a+1)bc}\:\stackrel{?}{\leqslant}\:1\,.$$

Taking advantage of $(3)$ we can write $${a\over a+1}\cdot a+{b\over b+1}\cdot b+{c\over c+1}\cdot c \;<\; a+b+c+2 \;=\; abc$$ and dividing by $\,abc\,$ gives the claim. This completes the proof.

Concluding remark
The upper bound of $\sqrt2$ on the RHS of the inequality is never attained, I'd guess. One gets close by via sending one variable to zero, say $a$, and keeping equal the remaining ones. Then $b=c=1+{2\over a}$ and the LHS evaluates to $${\sqrt{a+2}\over a+1}\:+\:{a^2\over (a+1)(a+2)}\,.$$

  • $\begingroup$ Bravo! The last inequality can be checked easier if you divide it by (a+1)(b+1)(c+1) $\endgroup$ – Minz Mar 25 '18 at 1:43

The hint.

After substitution $a=\frac{y+z}{x}$, $b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$, where $x$, $y$ and $z$ are positives,
and squaring of the both sides use AM-GM.

Indeed, let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y}$, where $x$, $y$ and $z$ are positives.

Thus, the condition gives $$\frac{1}{\frac{y+z}{x}+1}+\frac{1}{\frac{x+z}{y}+1}+\frac{1}{c+1}=1$$ or $$\frac{x+y}{x+y+z}+\frac{1}{c+1}=1$$ or $$\frac{1}{c+1}=\frac{z}{x+y+z}$$ or $$\frac{1}{c+1}=\frac{1}{\frac{x+y}{z}+1},$$ which gives $c=\frac{x+y}{z}$ and we need to prove that $$\sum_{cyc}\frac{\frac{y+z}{x}}{\sqrt{\frac{(x+y)(x+z)}{yz}}\left(\frac{y+z}{x}+1\right)}\leq\sqrt2$$ or $$\sum_{cyc}\frac{(y+z)\sqrt{yz}}{\sqrt{(x+y)(x+z)}}\leq\sqrt2(x+y+z)$$ or $$\sum_{cyc}\sqrt{yz(y+z)^3}\leq(x+y+z)\sqrt{2(x+y)(x+z)(y+z)}$$ or $$\sum_{cyc}\left(xy(x+y)^3+2x(x+y)(x+z)\sqrt{(x+y)(x+z)yz}\right)\leq2(x+y+z)^2\prod_{cyc}(x+y)$$ and since by AM-GM $$2\sqrt{(x+y)(x+z)yz}=2\sqrt{(xz+yz)(xy+yz)}\leq xz+yz+xy+yz=xy+xz+2yz,$$ it's enough to prove that $$\sum_{cyc}\left(xy(x+y)^3+x(x+y)(x+z)(xy+xz+2yz)\right)\leq2(x+y+z)^2\prod_{cyc}(x+y),$$ which is $$\sum_{cyc}(x^3y^2+x^3z^2+4x^3yz+6x^2y^2z)\geq0.$$ Done!

  • $\begingroup$ Cool, but I am confused that the Jacobian $\frac{\partial (a,b,c)}{\partial (x,y,z)}=0$ $\endgroup$ – Minz Mar 19 '18 at 13:52
  • $\begingroup$ @Minz If you want I am ready to write a full solution. It's a very easy now. $\endgroup$ – Michael Rozenberg Mar 19 '18 at 15:02
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    $\begingroup$ @Minz The proposed substitution is not a change of variables but rather a parametrisation of the constraint $\,1=\sum_{\text{cyc}}{1\over a+1}$, it is fulfilled for all choices of $x,y,z>0$. This implies that the Jacobian must be zero. $\endgroup$ – Hanno Mar 19 '18 at 16:33
  • $\begingroup$ @Hanno There are too many variables for parameterization, so they must be connected. For instance we may asume that x+y+z=1. $\endgroup$ – Minz Mar 20 '18 at 12:57
  • $\begingroup$ But this agreement returns me exactly at the very beginning (a+1=1/x, b+1=1/y, c+1=1/z) of my ugly solution a good continuation of which I did not find $\endgroup$ – Minz Mar 20 '18 at 13:08

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