For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$: $\sum_{cyc}\frac{4}{a+3b}\geq \sum_{cyc}\frac{3}{a+2b}$

Question

For $a,b,c\in\left[\frac{1}{\sqrt{6}}, 6\right]$ prove that $$\frac{4}{a+3b}+\frac{4}{b+3c}+\frac{4}{c+3a}\geq\frac{3}{a+2b}+\frac{3}{b+2c}+\frac{3}{c+2a}.$$

I can't really find a way to exploit the given condition. I noticed we can substitute $a\leftarrow \frac{\sqrt{6}}{a}$, $b\leftarrow \frac{\sqrt{6}}{b}$, $c\leftarrow \frac{\sqrt{6}}{c}$ preserving the conditions on $a,b,c$ which leads us to the same inequality for $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ where the terms are a bunch of weighted harmonic means.

Any ideas would be apreciated.

Answer 1

Since $6\sqrt6<15$ and our inequality is homogeneous,

it's enough to prove our inequality after following substitution.

Let $a=\frac{x+15}{x+1},$ $b=\frac{y+15}{y+1}$ and $c=\frac{z+15}{z+1},$ where $x$, $y$ and $z$ are non-negatives.

Thus, we need to prove that a sum of the following homogeneous polynomials is non-negative:

The homogeneous polynomial of the second degree it's $$202500\sum_{cyc}(x^2-xy)\geq0;$$ The homogeneous polynomial of the third degree it's: $$25\sum_{cyc}(12420x^3-783x^2y+5643x^2z-17280xyz)\geq0.$$ The homogeneous polynomial of the fourth degree it's: $$30\sum_{cyc}(3600x^4+10951x^3y+18329x^3z-2220x^2y^2-30660x^2yz)\geq0.$$ The homogeneous polynomial of the fifth degree it's: $$\sum_{cyc}(137373x^4y+214527x^4z+53095x^3y^2+248885x^3z^2+432000x^3yz-1085880x^2y^2z)\geq0.$$ The homogeneous polynomial of the sixth degree it's: $$\sum_{cyc}(33574x^4y^2+118106x^4z^2+30660x^3y^3+258300x^4yz-63260x^3y^2z+34460x^3z^2y-411840x^2y^2z^2)\geq0.$$ The homogeneous polynomial of the seventh degree: $$\sum_{cyc}(-3551x^4y^3+20291x^4z^3+73003x^4y^2z+108857x^4z^2y+1920x^3y^3z-200520x^3y^2z^2),$$ which can be negative.

The homogeneous polynomial of the eighth degree it's: $$\sum_{cyc}(480x^4y^4+5074x^4y^3z+13166x^4z^3y+15720x^4y^2z^2-34440x^3y^3z^2)\geq0.$$ The homogeneous polynomial of the ninth degree it's: $$xyz\sum_{cyc}(540x^3y^3+333x^3y^2z+1047x^3z^2y-1920x^2y^2z^2)\geq0.$$ The homogeneous polynomial of the tenth degree it's: $$60x^2y^2z^2\sum_{cyc}(x^2y^2-x^2yz)\geq0.$$ We see that it's enough to prove that: $$\sum_{cyc}(480x^4y^4+5074x^4y^3z+13166x^4z^3y+15720x^4y^2z^2-34440x^3y^3z^2)+$$ $$+\sum_{cyc}(-3551x^4y^3+20291x^4z^3+73003x^4y^2z+108857x^4z^2y+1920x^3y^3z-200520x^3y^2z^2)+$$ $$+\sum_{cyc}(33574x^4y^2+118106x^4z^2+30660x^3y^3+258300x^4yz-63260x^3y^2z+34460x^3z^2y-411840x^2y^2z^2)\geq0$$ or $$\sum_{cyc}(5074x^4y^3z+13166x^4z^3y+16200x^4y^2z^2-34440x^3y^3z^2)+$$ $$+\sum_{cyc}(16740x^4z^3+73003x^4y^2z+108857x^4z^2y+1920x^3y^3z-200520x^3y^2z^2)+$$ $$+\sum_{cyc}(84532x^4z^2+30660x^3y^3+258300x^4yz-63260x^3y^2z+34460x^3z^2y-344692x^2y^2z^2)+$$ $$+33574\sum_{cyc}(x^4y^2+x^4z^2-2x^2y^2z^2)-3551\sum_{cyc}(x^4y^3-x^4z^3)+480\sum_{cyc}(x^4y^4-x^4y^2z^2)\geq0,$$ for which it's enough to prove that: $$33574\sum_{cyc}(x^4y^2+x^4z^2-2x^2y^2z^2)-3551\sum_{cyc}(x^4y^3-x^4z^3)+480\sum_{cyc}(x^4y^4-x^4y^2z^2)\geq0.$$ Indeed, by AM-GM, C-S and Rearrangement we obtain: $$480\sum_{cyc}(x^4y^4-x^4y^2z^2)+33574\sum_{cyc}(x^4y^2+x^4z^2-2x^2y^2z^2)\geq$$ $$\geq2\sqrt{480\cdot33574\sum_{cyc}(x^4y^4-x^4y^2z^2)\sum_{cyc}(x^4y^2+x^4z^2-2x^2y^2z^2)}=$$ $$=2\sqrt{240\cdot33574\sum_{cyc}z^4(x^2-y^2)^2\sum_{cyc}z^2(x^2-y^2)^2}\geq$$ $$\geq3551\sqrt{\sum_{cyc}z^4(x^2-y^2)^2\sum_{cyc}z^2(x^2-y^2)^2}\geq$$ $$\geq3551\sum_{cyc}z^3(x^2-y^2)^2\geq3551\sum_{cyc}(x^4y^3-x^4z^3)$$ and we are done!

Answer 2

Ok this is P4 Seniors from Romania NMO 2015 SHL. My proof is:

$$\sum \frac{4}{a+3b} \geq \sum \frac{3}{a+2b} $$ $$\sum \left(\frac{4}{a+3b}- \frac{3}{a+2b}\right) \geq 0 $$ $$\sum \left(\frac{a-b}{(a+3b)(a+2b)}\right) \geq 0 $$ $$\sum \left(\frac{a-b}{(a+3b)(a+2b)} + \frac{a-b}{12ab}\right) \geq 0 $$ $$\sum \frac{(a-b)(-a^2+7ab+6b^2)}{(a+3b)(a+2b)}\geq 0 $$ $$\sum \frac{(a-b)^2(6b-a)}{(a+3b)(a+2b)} \geq 0 $$

which is obvious

Answer 3

New proof by Buffalo Way (BW).

Since the inequality is homogeneous, it suffices to prove the inequality for $a, b, c \in [1, 6\sqrt{6}]$. Since $6\sqrt{6} < 15$, it suffices to prove the inequality for $a, b, c \in [1, 15)$. After clearing the denominators, it suffices to prove that, for all $a, b, c \in [1, 15)$, \begin{align*} &6\,{a}^{4}b-6\,{a}^{4}c+35\,{a}^{3}{b}^{2}+25\,{a}^{3}{c}^{2}+25\,{a}^ {2}{b}^{3}-60\,{a}^{2}{b}^{2}c-60\,{a}^{2}b{c}^{2}\\ &+35\,{a}^{2}{c}^{3}- 6\,a{b}^{4}-60\,a{b}^{2}{c}^{2}+6\,a{c}^{4}+6\,{b}^{4}c+35\,{b}^{3}{c} ^{2}+25\,{b}^{2}{c}^{3}-6\,b{c}^{4}\\ \ge{}& 0. \tag{1} \end{align*}

Since the inequality is cyclic, assume that $c = \min(a, b, c)$. We split into two cases.

Case 1. $a \ge b \ge c$

Let $b = c + u, a = c + u + v$ for $u, v \ge 0$. (1) is written as \begin{align*} &120\,{c}^{3}{s}^{2}+120\,{c}^{3}st+120\,{c}^{3}{t}^{2}+300\,{c}^{2}{s} ^{3}+501\,{c}^{2}{s}^{2}t+321\,{c}^{2}s{t}^{2}\\ &\quad +60\,{c}^{2}{t}^{3}+240 \,c{s}^{4}+548\,c{s}^{3}t+402\,c{s}^{2}{t}^{2}+94\,cs{t}^{3}+60\,{s}^{ 5}\\ &\quad +173\,{s}^{4}t+166\,{s}^{3}{t}^{2}+59\,{s}^{2}{t}^{3}+6\,s{t}^{4}\\ \ge{}& 0 \end{align*} which is clearly true.

Case 2. $b \ge a \ge c$

Let $a = c + s, b = c + s + t$ for $s, t \ge 0$. (1) is written as \begin{align*} &\left( 120\,{s}^{2}+120\,st+120\,{t}^{2} \right) {c}^{3}+ \left( 300 \,{s}^{3}+399\,{s}^{2}t+219\,s{t}^{2}+60\,{t}^{3} \right) {c}^{2}\\ &\quad + \left( 240\,{s}^{4}+412\,{s}^{3}t+198\,{s}^{2}{t}^{2}+26\,s{t}^{3} \right) c\\ &\quad +60\,{s}^{5}+127\,{s}^{4}t+74\,{s}^{3}{t}^{2}+{s}^{2}{t}^{3} -6\,s{t}^{4}\\ \ge{}& 0. \tag{2} \end{align*} Since $c\ge 1$, it suffices to prove (2) when $c = 1$, i.e., \begin{align*} &60\,{s}^{5}+127\,{s}^{4}t+74\,{s}^{3}{t}^{2}+{s}^{2}{t}^{3}-6\,s{t}^{4 }+240\,{s}^{4}+412\,{s}^{3}t\\ &\quad +198\,{s}^{2}{t}^{2}+26\,s{t}^{3}+300\,{s} ^{3}+399\,{s}^{2}t+219\,s{t}^{2}+60\,{t}^{3}\\ &\quad +120\,{s}^{2}+120\,st+120 \,{t}^{2}\\ \ge{}& 0. \tag{3} \end{align*} From $b = c + s + t < 15$, we have $s + t < 15$ and thus $s, t < 15$. It suffices to prove (3) for all $s, t \in [0, 15)$. Let $s = \frac{15u}{1 + u}, t = \frac{15v}{1 + v}$ for $u, v \ge 0$. (3) is equivalently written as \begin{align*} &\left( 71742\,{u}^{5}+86649\,{u}^{4}+30218\,{u}^{3}+1350\,{u}^{2}-393 \,u+68 \right) {v}^{4}\\ &\quad + \left( 215487\,{u}^{5}+232175\,{u}^{4}+81094\, {u}^{3}+11562\,{u}^{2}+1232\,u+76 \right) {v}^{3}\\ &\quad + \left( 229781\,{u}^ {5}+194962\,{u}^{4}+51173\,{u}^{3}+5267\,{u}^{2}+283\,u+8 \right) {v}^ {2}\\ &\quad + \left( 104794\,{u}^{5}+59060\,{u}^{4}+8721\,{u}^{3}+463\,{u}^{2}+ 8\,u \right) v\\ &\quad +17408\,{u}^{5}+4224\,{u}^{4}+324\,{u}^{3}+8\,{u}^{2} \\ \ge{}&0 \end{align*} which is true since all the coefficients (as a polynomial in $v$) are non-negative. (Note: The coefficient of $v^4$ is $71742\,{u}^{5}+86649\,{u}^{4}+30218\,{u}^{3}+1350\,{u}^{2}-393 \,u+68\ge 1350\,{u}^{2}-393 \,u+68 \ge 0$.)

We are done.