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The inequality is $$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$ for $x,y,z,t>0$.

It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).

The three variables simpler case $$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$ can be proved using the Cauchy-Schwarz inequality in "Engel's form".

I have tried variants of Holder type inequalities, until now unsuccessfully.

Michael Rozenberg
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JohnnyC
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3 Answers3

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This is a solution using Carlson inequality: if $x_{ij}(1\leq i\leq m,1\leq j\leq n)$ are non-negative reals, then $$ \prod_{s=1}^m\sum_{t=1}^nx_{st}\geq\left(\sum_{t=1}^n\left(\prod_{s=1}^mx_{st}\right)^{\frac{1}{m}}\right)^{m} $$

Proof. According to the Carlson inequality, $$ \begin{align} &\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot\sum(x+y)\cdot\sum(x+z)(x+t)\\ \geq&\left(\sum x\right)^3\\ \because&\sum(x+y)=2(x+y+z+t),\sum(x+z)(x+t)=(x+y+z+t)^2\\ \therefore&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot2(x+y+z+t)\cdot(x+y+z+t)^2\geq(x+y+z+t)^3 \end{align} $$ One can now easily observe that the original inequality is true.
atzlt
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As inequality homogenous WLOG $x+y+z+t=4$

By AM-GM$$\sum \frac{x^3}{(x+y)(x+t)(x+z)} \ge\sum \frac{27x^3}{{(3x+y+z+t)}^3}= \sum \frac{27x^3}{{(2x+4)}^3}$$ By tangent line method $$\frac{x^{3}}{\left(2x+4\right)^{3}}\ge \frac{1}{6^{3}}+\frac{1}{108}\left(x-1\right)$$ $$\iff -\frac{\left(x-1\right)^{2}\left(x^{2}-6x-4\right)}{108\left(x+2\right)^{3}}\ge 0$$ which is true as $x^2-6x-4<0,\forall x\in[0,4]$ So $$ \sum \frac{27x^3}{{(2x+4)}^3}\ge \frac{27 \cdot 4}{6^3}+\frac{0}{108}=\frac{1}{2}$$ done!!

Albus Dumbledore
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  • I think $\frac{x^3}{(2x+4)^3}$ should be convex on $(0,4]$ in order for the inequality by tangent line method to hold. – JohnnyC Feb 15 '21 at 17:33
  • @JohnnyC no that is not required ,by the way I just proved that inequality by some simple factorisation in the next line..do you see it? – Albus Dumbledore Feb 16 '21 at 02:07
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    @AlbusDumbledore Nice TL method. (+1) – River Li Feb 16 '21 at 03:20
  • @RiverLi thank you! – Albus Dumbledore Feb 16 '21 at 03:53
  • @AlbusDumbledore: You're right; sorry. The function $\frac{x^3}{(2x+4)^3}$ changes convexity in $x=2$, but it continues to steep up, so the tangent in $x=1$ remains under the graph on the interval. Your solution hints to a technique that can be used for many other (hard) inequalities. Thanks! – JohnnyC Feb 16 '21 at 08:44
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We need to prove that: $$2\sum_{cyc}x^3(y+z)(y+t)(z+t)\geq(x+y)(x+z)(x+t)(y+z)(y+t)(z+t)$$ or $$2\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt\right)\geq\sum_{sym}\left(x^3y^2z+\frac{1}{3}x^3yzt+x^2y^2zt+\frac{1}{3}x^2y^2z^2\right)$$ or $$\sum_{sym}(3x^3y^2z-x^2y^2z^2+x^3yzt-3x^2y^2zt)\geq0,$$ which is true by Muirhead because

$(3,2,1,0)\succ(2,2,2,0),$ $(3,2,1,0)\succ(2,2,1,1)$ and $(3,1,1,1)\succ(2,2,1,1).$

Michael Rozenberg
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