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

Question

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.

Answer 1

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 $$u^4+4u^4+6u^2-8u-20=u^4+4u^2(u-2)+4u(u-2)+10(u^2-2)\ge 0$$

Answer 2

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.

Answer 3

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!