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Let $a, b, c, d > 0$. Prove that $$\frac{a}{b^2+c^2+d^2}+\frac{b}{a^2+c^2+d^2}+\frac{c}{a^2+b^2+d^2}+\frac{d}{a^2+b^2+c^2}$$ $$\geq\frac{3\sqrt{3}}{2}\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}.$$

What I tried, was to say that $a^2+b^2+c^2+d^2=1$ and so $$\frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2}+\frac{d}{1-d^2}≥\frac{3\sqrt{3}}{2}$$ and $$-\frac{a}{(a+1)(a-1)}-\frac{b}{(b+1)(b-1)}-\frac{c}{(c+1)(c-1)}-\frac{d}{(d+1)(d-1)}≥\frac{3\sqrt{3}}{2}$$ and so $$\frac{a}{a+1}-\frac{a}{a-1}+\frac{b}{b+1}-\frac{b}{b-1}+\frac{c}{c+1}-\frac{c}{c-1}+\frac{d}{d+1}-\frac{d}{d-1}≥\frac{3\sqrt{3}}{2}$$ Perhaps this leads somewhere?

River Li
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Severus156
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2 Answers2

2

Since the desired inequality is homogeneous, assume that $a^2 + b^2 + c^2 + d^2 = 1$. The desired inequality becomes $$\sum_{\mathrm{cyc}}\frac{a}{1 - a^2} \ge \frac{3\sqrt 3}{2}.$$

Note that \begin{align*} \frac{a}{1 - a^2} - \frac{3\sqrt 3}{2}a^2 &= \frac{a}{2(1 - a^2)} [2 - 3\sqrt 3\, a(1 - a^2)]\\ &= \frac{a}{2(1 - a^2)}\frac{4 - 27a^2(1 - a^2)^2}{2 + 3\sqrt 3\, a(1 - a^2)}\\ &\ge 0 \end{align*} where we have used AM-GM to get $2a^2(1 - a^2)^2 \le \left(\frac{2a^2 + 1-a^2 + 1 - a^2}{3}\right)^3 = \frac{8}{27}$.

Thus, we have $$\sum_{\mathrm{cyc}} \frac{a}{1 - a^2} \ge \frac{3\sqrt 3}{2}(a^2 + b^2 + c^2 + d^2) = \frac{3\sqrt 3}{2}.$$

We are done.

River Li
  • 49,125
1

It's wrong! Try $a=b=c=d=-1$.

For positive variables it's true.

Indeed, by Holder we obtain: $$\left(\sum_{cyc}\frac{a}{b^2+c^2+d^2}\right)^2\sum_{cyc}a^2(b^2+c^2+d^2)^2\geq(a^2+b^2+c^2+d^2)^3.$$ Let $a^2=x$, $b^2=y$, $c^2=z$ and $d^2=t$.

Thus, it's enough to prove that $$4(x+y+z+t)^4\geq27\sum_{cyc}x^2(y+z+t)^2.$$ Now, let $x=\min\{x,y,z,t\},$ $y=x+u,$ $z=x+v$ and $t=x+w$.

Thus, $u$, $v$ and $w$ are non-negatives and $$4(x+y+z+t)^4-27\sum_{cyc}x^2(y+z+t)^2=$$ $$=52x^4+52(u+v+w)x^3+12\sum_{cyc}(5u^2+uv)x^2+$$ $$+4\sum_{cyc}(16u^3-6u^2v-6u^2w+5uvw)x+$$ $$+2\sum_{cyc}(2u^4+8u^3v+8u^3w-15u^2v^2-3u^2vw)\geq0$$ because by Muirhead $$\sum_{cyc}(16u^3-6u^2v-6u^2w+5uvw)\geq \sum_{cyc}(12u^3-6u^2v-6u^2w)\geq0$$ and $$\sum_{cyc}(2u^4+8u^3v+8u^3w-15u^2v^2-3u^2vw)\geq0.$$ Done!

The inequality $$4(x+y+z+t)^4\geq27\sum_{cyc}x^2(y+z+t)^2$$ we can prove also by the Lagrange multipliers method.

Indeed let $x+y+z+t=3$, where $x$, $y$, $z$ and $t$ be non-negatives.

Thus, we need to prove that $$\sum_{cyc}x^2(3-x)^2\leq12.$$ Now, let $$f(x,y,z,t)=\sum_{cyc}x^2(3-x)^2+\lambda(x+y+z+t-3).$$ Thus, $f$ is a continuous function on the compact $$C=\{(x,y,z,t)|x\geq0,y\geq0,z\geq0,t\geq0,x+y+z+t=3\},$$ which says that $f$ gets on $C$ the maximal value.

Let $(x,y,z,t)$ be a maximum point.

In this point we have two cases:

  1. One of our variables is equal to zero.

Let $t=0$.

Thus, we need to prove that $$\sum_{cyc}x^2(3-x)\leq12, $$ where $x+y+z=3$ or $$\sum_{cyc}(4-x^2(3-x)^2)\geq0$$ or $$\sum_{cyc}(4-x^2(3-x)^2+4(x-1))\geq0$$ or $$\sum_{cyc}x(4-x)(x-1)^2\geq0,$$ which is obvious.

  1. $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}=\frac{\partial f}{\partial t}=0,$$ which gives $$9x-9x^2+2x^3=9y-9y^2+2y^3=9z-9z^2+2z^3=9t-9t^2+2t^3$$ and from here we obtain: $$(x-y)(2x^2+2xy+2y^2-9x-9y+9)=0,$$ $$(x-z)(2x^2+2xz+2z^2-9x-9z+9)=0,$$ $$(x-t)(2x^2+2xt+2t^2-9x-9t+9)=0,$$ $$(y-z)(2y^2+2yz+2z^2-9y-9z+9)=0,$$ $$(y-t)(2y^2+2yt+2t^2-9y-9t+9)=0$$ and $$(z-t)(2z^2+2zt+2t^2-9z-9t+9)=0.$$ Now, let $x\neq y$, $x\neq z$ and $x\neq t.$

Thus, $$2x^2+2xy+2y^2-9x-9y+9=2x^2+2xz+2z^2-9x-9z+9=2x^2+2xt+2t^2-9x-9t+9$$ or $$(y-z)(2(x+y+z)-9)=0$$ and $$(y-t)(2(x+y+t)-9=0$$ or $$(y-z)(2(3-t)-9)=0$$ and $$(y-t)(2(3-z)-9)=0,$$ which gives $y=z=t.$

If $x=y$, but $y\neq z$ and $y\neq t$ we obtain $$2y^2+2yz+2z^2-9y-9z+9=2y^2+2yt+2t^2-9y-9t+9$$ or $$(z-t)(2x+3)=0,$$ which gives $z=t$.

Id est, it's enough to prove $\sum\limits_{cyc}x^2(3-x)\leq12$ in two cases:

  1. $y=z=x$ and $t=3-3x,$ where $0\leq x\leq1$.

Thus, we need to prove that $$3x^2(3-x)^2+9(3-3x)^2x^2\leq12$$ or $$(1-x)(7x^3-8x^2+x+1)\geq0,$$ which is true because by AM-GM $$7x^3-8x^2+x+1=3\cdot\frac{7}{3}x^3+x+1-8x^2\geq$$ $$\geq5\sqrt[5]{\left(\frac{7}{3}x\right)^3\cdot x\cdot1}-8x^2=\left(5\sqrt[5]{\frac{343}{27}}-8\right)x^2\geq0.$$ 2. $x=y$, $z=t$.

Thus, $z=t=\frac{3}{2}-x,$ where $0\leq x\leq\frac{3}{2}$ and we need to prove that $$2x^2(3-x)^2+2\left(\frac{3}{2}-x\right)^2\left(\frac{3}{2}+x\right)^2\leq12$$ or $$x^2(3-2x)^2\leq\frac{15}{8}$$ or $$x(3-2x)\leq\sqrt{\frac{15}{8}},$$ which is true by AM-GM again: $$x(3-2x)=\frac{1}{2}2x(3-2x)\leq\frac{1}{2}\left(\frac{2x+3-2x}{2}\right)^2=\frac{9}{8}<\sqrt{\frac{15}{8}}.$$

Michael Rozenberg
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