# Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$

The inequality:

$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$ Conditions: $a,b,c,d \in \mathbb{R^+}$

I tried using the normal Cauchy-Scharwz, AM-RMS, and all such.. I think I can do it using some bash method like expanding the LHS and then maybe to go with homogeneity, normalization and then Muirhead, Jensen or something I dont know since I didn't go that way..

But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.

• Any conditions etc. mentioned on $a,b,c,d$?(quite expected)... Commented Jan 12, 2017 at 17:58
• All are positive reals Commented Jan 12, 2017 at 17:59
• it can be proved with BW! Commented Jan 12, 2017 at 18:06
• @Dr.SonnhardGraubner What is BW? Can you please provide a solution? Thanks in advance Commented Jan 12, 2017 at 18:07
• Oh do please send in your solution. Thanks Commented Jan 12, 2017 at 18:09

By Holder inequality,we have $$\left(\sum\dfrac{a^2}{(a+b+c)^2}\right)\left(\sum a(a+b+c)\right)^2\ge \left(\sum_{cyc} a^{\frac{4}{3}}\right)^3$$ therefore $$\sum\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{(a^{4/3}+b^{4/3}+c^{4/3}+d^{4/3})^3}{[(a+c)^2+(b+d)^2+(a+c)(b+d)]^2}$$ use Holder we have $$a^{\frac{4}{3}}+c^{\frac{4}{3}}\ge 2\left(\dfrac{a+c}{2}\right)^{\frac{4}{3}}$$ $$b^{\frac{4}{3}}+d^{\frac{4}{3}}\ge 2\left(\dfrac{b+d}{2}\right)^{\frac{4}{3}}$$if setting $$t^3=\dfrac{(a+c)}{b+d}$$,we have $$\sum_{cyc}\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{1}{2}\cdot\dfrac{(t^4+1)^3}{(t^6+t^3+1)^2}$$ it remain to show that $$9(t^4+1)^3\ge 8(t^6+t^2+1)^2,t>0$$ or $$9\left(t^2+\dfrac{1}{t^2}\right)^3\ge 8\left(t^3+\dfrac{1}{t^3}+1\right)^2$$ let $u=t+\dfrac{1}{t}\ge 2$ the above ineuqality becomes $$9(u^2-2)^2\ge 8(u^3-3u+1)^2$$ or $$(u-2)^2(u^4+4u^3+6u^2-8u-20)\ge 0$$ since $$u^4+4u^4+6u^2-8u-20=u^4+4u^2(u-2)+4u(u-2)+10(u^2-2)\ge 0$$

Trying the Buffalo Way suggested in the comments.

Assume without loss of generality that $a\le b\le c\le d$.

Let $a=x_1$, $b=x_1+x_2$, $c=x_1+x_2+x_3$, $d=x_1+x_2+x_3+x_4$. So $x_i\ge0,i\in[1,4]$

When then want to consider the sign of:

$$\left(\frac{x_1}{3x_1+2x_2+x_3}\right)^2+\left(\frac{x_1+x_2}{3x_1+3x_2+2x_3+x_4}\right)^2+\left(\frac{x_1+x_2+x_3}{3x_1+2x_2+2x_3+x_4}\right)^2+\left(\frac{x_1+x_2+x_3+x_4}{3x_1+2x_2+x_3+x_4}\right)^2-\frac{4}{9}$$

Using brute force gives a disgusting fraction with denominator of:

$$9 (3 x_1+2 x_2+x_3)^2 (3 x_1+2x_2+x_3+x_4)^2 (3 x_1+2 x_2+2x_3+x_4)^2 (3 x_1+3 x_2+2 x_3+x_4)^2$$

and has156 terms in the numerator:

$$4374x_1^6x_2^2+20412x_1^5x_2^3+39366x_1^4x_2^4+40176x_1^3x_2^5+22896x_1^2x_2^6+6912x_1x_2^7+864x_2^8+8748x_1^6x_2x_3+53946x_1^5x_2^2x_3+129600x_1^4x_2^3x_3+158598x_1^3x_2^4x_3+105516x_1^2x_2^5x_3+36456x_1x_2^6x_3+5136x_2^7x_3+8748x_1^6x_3^2+65610x_1^5x_2x_3^2+191727x_1^4x_2^2x_3^2+280260x_1^3x_2^3x_3^2+218367x_1^2x_2^4x_3^2+86820x_1x_2^5x_3^2+13868x_2^6x_3^2+26244x_1^5x_3^3+131058x_1^4x_2x_3^3+258876x_1^3x_2^2x_3^3+250614x_1^2x_2^3x_3^3+118584x_1x_2^4x_3^3+21944x_2^5x_3^3+32481x_1^4x_3^4+118908x_1^3x_2x_3^4+162774x_1^2x_2^2x_3^4+98148x_1x_2^3x_3^4+21925x_2^4x_3^4+21222x_1^3x_3^5+55566x_1^2x_2x_3^5+48414x_1x_2^2x_3^5+13982x_2^3x_3^5+7713x_1^2x_3^6+13050x_1x_2x_3^6+5509x_2^2x_3^6+1476x_1x_3^7+1220x_2x_3^7+116x_3^8+7290x_1^5x_2^2x_4+28350x_1^4x_2^3x_4+43740x_1^3x_2^4x_4+33480x_1^2x_2^5x_4+12720x_1x_2^6x_4+1920x_2^7x_4+8748x_1^6x_3x_4+52488x_1^5x_2x_3x_4+143370x_1^4x_2^2x_3x_4+209952x_1^3x_2^3x_3x_4+167958x_1^2x_2^4x_3x_4+69000x_1x_2^5x_3x_4+11384x_2^6x_3x_4+39366x_1^5x_3^2x_4+180306x_1^4x_2x_3^2x_4+341172x_1^3x_2^2x_3^2x_4+325512x_1^2x_2^3x_3^2x_4+154296x_1x_2^4x_3^2x_4+28848x_2^5x_3^2x_4+64476x_1^4x_3^3x_4+225504x_1^3x_2x_3^3x_4+300762x_1^2x_2^2x_3^3x_4+179304x_1x_2^3x_3^3x_4+39998x_2^4x_3^3x_4+52002x_1^3x_3^4x_4+132408x_1^2x_2x_3^4x_4+113484x_1x_2^2x_3^4x_4+32534x_2^3x_3^4x_4+22302x_1^2x_3^5x_4+37044x_1x_2x_3^5x_4+15474x_2^2x_3^5x_4+4878x_1x_3^6x_4+3982x_2x_3^6x_4+428x_3^7x_4+4374x_1^6x_4^2+18954x_1^5x_2x_4^2+39123x_1^4x_2^2x_4^2+48276x_1^3x_2^3x_4^2+35658x_1^2x_2^4x_4^2+14280x_1x_2^5x_4^2+2360x_2^6x_4^2+24786x_1^5x_3x_4^2+98172x_1^4x_2x_3x_4^2+166212x_1^3x_2^2x_3x_4^2+147888x_1^2x_2^3x_3x_4^2+67668x_1x_2^4x_3x_4^2+12504x_2^5x_3x_4^2+51759x_1^4x_3^2x_4^2+167184x_1^3x_2x_3^2x_4^2+210438x_1^2x_2^2x_3^2x_4^2+121080x_1x_2^3x_3^2x_4^2+26558x_2^4x_3^2x_4^2+51192x_1^3x_3^3x_4^2+124074x_1^2x_2x_3^3x_4^2+102816x_1x_2^2x_3^3x_4^2+28928x_2^3x_3^3x_4^2+25911x_1^2x_3^4x_4^2+41676x_1x_2x_3^4x_4^2+17072x_2^2x_3^4x_4^2+6486x_1x_3^5x_4^2+5190x_2x_3^5x_4^2+637x_3^6x_4^2+5832x_1^5x_4^3+20574x_1^4x_2x_4^3+30780x_1^3x_2^2x_4^3+24642x_1^2x_2^3x_4^3+10488x_1x_2^4x_4^3+1864x_2^5x_4^3+19764x_1^4x_3x_4^3+58320x_1^3x_2x_3x_4^3+67752x_1^2x_2^2x_3x_4^3+36756x_1x_2^3x_3x_4^3+7792x_2^4x_3x_4^3+25596x_1^3x_3^2x_4^3+57996x_1^2x_2x_3^2x_4^3+45636x_1x_2^2x_3^2x_4^3+12442x_2^3x_3^2x_4^3+15606x_1^2x_3^3x_4^3+23916x_1x_2x_3^3x_4^3+9506x_2^2x_3^3x_4^3+4488x_1x_3^4x_4^3+3488x_2x_3^4x_4^3+494x_3^5x_4^3+2916x_1^4x_4^4+7938x_1^3x_2x_4^4+8469x_1^2x_2^2x_4^4+4272x_1x_2^3x_4^4+864x_2^4x_4^4+6480x_1^3x_3x_4^4+13572x_1^2x_2x_3x_4^4+9984x_1x_2^2x_3x_4^4+2608x_2^3x_3x_4^4+5148x_1^2x_3^2x_4^4+7380x_1x_2x_3^2x_4^4+2816x_2^2x_3^2x_4^4+1728x_1x_3^3x_4^4+1292x_2x_3^3x_4^4+214x_3^4x_4^4+648x_1^3x_4^5+1242x_1^2x_2x_4^5+840x_1x_2^2x_4^5+208x_2^3x_4^5+864x_1^2x_3x_4^5+1128x_1x_2x_3x_4^5+408x_2^2x_3x_4^5+354x_1x_3^2x_4^5+252x_2x_3^2x_4^5+50x_3^3x_4^5+54x_1^2x_4^6+60x_1x_2x_4^6+20x_2^2x_4^6+30x_1x_3x_4^6+20x_2x_3x_4^6+5x_3^2x_4^6$$

All the co-efficients of the terms are positive and $x_i\ge0,i\in[1,4]$ so the entire expression is positive.

Reflection

Buffalo Way is terrible.

Also I don't know how to format that eyesore of a numerator nicely on this site.

• @mixedmath Is there a way to put mathjax inside a horizontally scrolling box? I couldn't work it out. Commented Jan 14, 2017 at 1:17
• No, there isn't a way. Commented Jan 14, 2017 at 3:22
• To the downvoter can you please indicate what is actually wrong with my calculations? I have acknowledged they are very inefficient and that it is a terrible technique (Blame the person who originally suggested it would work not me). I don't believe there is anything technically wrong with the calculations so if you did repeat them and find an error please let me know. Commented Jan 22, 2017 at 0:11

Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$, $x+y+z+t=4$, $x+t=2u$ and $y+z=2v$.

Hence, $u+v=2$ and by Rearrangement we obtain: $$\sum\limits_{cyc}\frac{a^2}{(a+b+c)^2}\geq\frac{x^2}{(x+y+z)^2}+\frac{y^2}{(x+y+t)^2}+\frac{z^2}{(x+z+t)^2}+\frac{t^2}{(y+z+t)^2}=$$ $$=\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}+\frac{y^2}{(4-z)^2}+\frac{z^2}{(4-y)^2}$$ and since $$\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}\geq\frac{-2(x+t)^2+16(x+t)-16}{(8-x-t)^2} \tag1$$ it's just $$((x+t)(x^2+t^2)-12(x^2+xt+t^2)+48(x+t)-64)^2\geq0, \tag2$$ it remains to prove that $$\frac{-2u^2+8u-4}{(4-u)^2}+\frac{-2v^2+8v-4}{(4-v)^2}\geq\frac{4}{9},\tag3$$ where $u$ and $v$ are positive numbers such that $u+v=2$ or $$\sum_{cyc}\left(\frac{-2u^2+8u-4}{(4-u)^2}-\frac{2}{9}\right)\geq0$$ or $$\sum\limits_{cyc}\frac{(u-1)(17-5u)}{(4-u)^2}\geq0$$ or $$\sum\limits_{cyc}\left(\frac{(u-1)(17-5u)}{(4-u)^2}-\frac{4}{3}(u-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(u-1)^2(13-4u)}{(4-u)^2}\geq0.$$ Done!

• Would you mind adding some explanation of what motivates the brilliant and insightful steps Equations (1), (2) and (3)? Could you also reveal your tricks for proving these inequalities without brute force? Thank you.
– Hans
Commented Aug 3, 2017 at 8:11
• @Hans The motivation is a clear enough: we want to reduce a number of variables and to make a sum of squares. But I remember releasing of this idea was very hard for me. I remember that I used brute force for the verification. Commented Aug 3, 2017 at 8:31