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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

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1 Answer 1

up vote 10 down vote accepted
+50

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.

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1  
He does not assume this. The rearrangement inequality only makes use of the fact that there is a way to arrange them in ascending order, might be a,b,c,d or a,c,d,b or whatever. However, there are two reasons why I don't like this solution: The last part is a really ugly calculation (isn't there another way of proving the lemma?) and we don't learn anything about the equality case. –  thomas Jan 4 at 18:30
    
Ah, you are right :) Now that's particularly nice to know and I'm wondering: Why isn't there some abstract nonsense argument (e.g. by symmetry) that immediately shows that the term is maximized iff all of the numbers agree? And why do we have the extra solutions in three dimensions? Does this inequality still hold in higher dimensional spheres? –  thomas Jan 4 at 18:52
    
Very much sustained! Let $a=b=c=d$ then $a^2+b^2+c^2+d^2=4$ is of course fulfilled if and only if $a=b=c=d=1$; nobody would be surprised if $f(a,b,c,d) = a/(b+3)+b/(c+3)+c/(d+3)+d/(a+3)$ assumes its one and only maximum $f(a,b,c,d)=1/4+1/4+1/4+1/4=1$ for these values of $(a,b,c,d)$ . It's frustrating that there seems not to exist a theorem somewhere that simply guarantees a maximum or a minimum of a function when all variables in a problem with such high symmetry are just equal. I mean: a formalization of "by symmetry". Why doesn't e.g. GROUP THEORY come in here? –  Han de Bruijn Jan 4 at 19:39
    
I think the fisrt is not true.for example $$x^2y+y^2z+z^2x\ge xy^2+yz^2+zx^2$$, then we have $a,b,c$ be a permutation of $x,y,z$ such $c\le b\le a$,then $$a^2b+b^2c+c^2a-(ab^2+bc^2+ca^2)=(a-b)(b-c)(a-c)\ge 0$$. in fact, $a=0,b=1,c=2$ this inequality is not true –  math110 Jan 5 at 12:23
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why $$12XY\ge 6X^2Y+6Y^2X \Longleftrightarrow 6XY(X^2+Y^2-2)\le 0$$but maybe $XY\le 0$ –  user94270 Jan 6 at 6:41

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