How prove this inequality $\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3}\le 1$

Question:

let $a,b,c,d\ge 0$,such $$a^2+b^2+c^2+d^2=4$$ show that $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$

My try: By Cauchy-Schwarz inequality,we have $$\sum_{cyc}\dfrac{a}{b+3}\le\sqrt{(\sum_{cyc}a^2)\left(\sum_{cyc}\dfrac{1}{(b+3)^2}\right)}$$ then we have only prove this $$\sum_{cyc}\dfrac{1}{(a+3)^2}\le \dfrac{1}{4}?$$ This is not true,in fact, we have $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge \dfrac{1}{4}?$$ because we have $$\dfrac{1}{(a+3)^2}\ge\dfrac{5-a^2}{64}$$ this is true because $$\Longleftrightarrow \dfrac{(a-1)^2(a^2+8a+19)}{64(a+3)^2}\ge 0$$ so $$\sum_{cyc}\dfrac{1}{(a+3)^2}\ge\sum_{cyc}\dfrac{5-a^2}{64}=\dfrac{1}{4}$$ can see:http://www.wolframalpha.com/input/?i=1%2F%28a%2B3%29%5E2-%285-a%5E2%29%2F64

This methods is from: can see:Prove this equality $\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}$

By the way I have see this three variable inequality

let $a,b,c$ be non-negative numbers such that $$a^2+b^2+c^2=3$$ show that $$\dfrac{a}{b+2}+\dfrac{b}{c+2}+\dfrac{c}{a+2}\le 1$$

proof: By expanding,the inequality becomes $$ab^2+bc^2+ca^2\le abc+2$$ without loss of generality,assume that

$$\min(a,b,c)\le b\max(a,b,c)$$ then \begin{align*} 2-ab^2-bc^2-ca^2+abc&=2-ab^2-b(3-a^2-b^2)-ca^2+abc\\ &=(b^3-3b+2)-a(b^2-ab+ca-bc)\\ &=(b-1)^2(b+2)-a(b-a)(b-c)\ge 0 \end{align*} Equality occurs for $(a,b,c)=(1,1,1)$ and also for $(a,b,c)=(0,1,\sqrt{2})$ or any cyclic permutation.

so my Four-inequality variable inequality,can use this methods $$\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{d+3}+\dfrac{d}{a+3}\le 1$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+9a^2+ab^2d+3ab^2+abc^2-abcd+3ac^2+9ac+3b^2d+9b^2+3bc^2+bcd^2+3bd^2+9bd+9c^2+3cd^2+9d^2-81\le 0$$ $$\Longleftrightarrow a^2cd+3a^2c+3a^2d+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow a^2(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$ $$\Longleftrightarrow (4-b^2-c^2-d^2)(cd+3c+3d)+ab^2d+3ab^2+abc^2+3ac^2+9ac+3b^2d+3bc^2+bcd^2+3bd^2+9bd+3cd^2\le 45$$

Then I can't

Thank you very much

Solution 1:

Let $a_1, a_2, a_3$ and $a_4$ be a permutation of $a, b, c$ and $d$ such that $a_1 \leqslant a_2 \leqslant a_3 \leqslant a_4.$ It follows from the rearrangement inequality that

$$\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}+\frac{d}{a+3} \leqslant \frac{a_1}{a_4+3}+\frac{a_2}{a_3+3}+\frac{a_3}{a_2+3}+\frac{a_4}{a_1+3}.$$

A bit later I'm going to prove that $\displaystyle \forall x,y \in [0,2] \;\; \frac{x}{y+3}+\frac{y}{x+3} \leqslant \frac{3}{32}x^2+\frac{3}{32}y^2+\frac{5}{16}.$

Assuming it's true your inequality can be solved pretty quickly:

$$\frac{a_1}{a_4+3}+\frac{a_2}{a_3+3}+\frac{a_3}{a_2+3}+\frac{a_4}{a_1+3} \leqslant \left(\frac{3}{32}a_1^2+\frac{3}{32}a_4^2+\frac{5}{16}\right) + \left(\frac{3}{32}a_2^2+\frac{3}{32}a_3^2+\frac{5}{16}\right) = \frac{3}{32}\bigl(a_1^2 + a_2^2 + a_3^2 + a_4^2\bigr)+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1$$

Lemma. $\displaystyle \frac{x}{y+3}+\frac{y}{x+3} \leqslant \frac{3}{32}x^2+\frac{3}{32}y^2+\frac{5}{16} \quad \forall x,y \in [0,2].$

Consider the substitution: $x = X+1,\; y = Y+1.$ Then $X,Y \in [-1, 1]$ and this inequality turns into:

$$\frac{X+1}{Y+4}+\frac{Y+1}{X+4} \leqslant \frac{3}{32}(X+1)^2+\frac{3}{32}(Y+1)^2+\frac{5}{16}$$ $$\Longleftrightarrow 3X^3Y + 3XY^3 + 18X^2Y + 18XY^2 +12X^3 + 12Y^3\\ + 40X^2 + 40Y^2 + 64XY \geqslant 0.$$

If $XY\geqslant0$:

$$3XY(X^2 + Y^2) + 18XY(X + Y) + 12(X^3 + Y^3) + 40(X^2 + Y^2) + 64XY = 9(X^2+Y^2)(X+1)(Y+1) + 3(X+Y+2)(X+Y)^2 + 20(X+Y)^2\\ + 5(X^2+Y^2) + 6XY(2-X-Y) \geqslant 0.$$

If $XY<0$ assuming WLOG that $X\in(0,1],\ Z=-Y\in(0,1]$ (thanks to this answer):

$$-3XZ(X^2 + Z^2) - 18XZ(X - Z) + 12(X^3 - Z^3) + 40(X^2 + Z^2) - 64XZ \geqslant 0\\ 18XZ^2 + 12X^3 + 40(X^2 + Z^2) \geqslant 3XZ(X^2 + Z^2) + 18X^2Z + 12Z^3 + 64XZ,$$

which follows from

$$\begin{split} 12Z^2 \geqslant& 12Z^3\\ 40X^2+28Z^2 \geqslant& 2\sqrt{40\cdot28}XZ \geqslant 64XZ\\ 9XZ^2+9X^3 \geqslant& 18X^2Z\\ 3X^3 \geqslant& 3X^3Z\\ 3XZ^2 \geqslant& 3XZ^3\\ 6XZ^2 \geqslant& 0. \end{split}$$

This ends the proof.