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Let $a$, $b$, $c$ and $d$ are non-negative numbers such that $abc+abd+acd+bcd=4.$ Prove that:

$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}\leq1$

I simplified it and it turns out that it suffices to show

$19\le3(ab+ac+ad+bc+bd+cd)+abcd$

subject to $abc+abd+acd+bcd=4.$

I then tried Lagrange multiplier but it didn't help too much...

Martin Sleziak
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user70520
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    Left side is continuous and convex in each variable a,b,c. I hope they are non negative reals. Apply the fact that if it is continuous and convex it reaches its maximum on the constraints. – Torsten Hĕrculĕ Cärlemän Jul 24 '13 at 09:26
  • Torsten: how to show that the function reaches its max on the constraints if it is continuous and convex in each variable? Thanks. – user70520 Jul 24 '13 at 09:28
  • Look at say, any one dimensional curve. If it is continuous and convex, then it doesnt have any bends or downward slopes. By convex, both the curve and its slope increase or decrease. In this case $a,b,c,d$ have no bounds. Only thing holding them is the constraint. So use it to find the extrema. – Torsten Hĕrculĕ Cärlemän Jul 24 '13 at 09:31
  • @AlbertH, I have deleted my post, however, I see the convexity method to be correct. Please correct me. – Torsten Hĕrculĕ Cärlemän Jul 24 '13 at 10:40

2 Answers2

1

first, we can easily prove if $a=0,bcd=4 \not = 0 $ and it is true for the inequality.(AM-GM, $19 \le 18 \sqrt[3] {2} \iff (18+1)^3 \le 2 \times 18^3 \iff 3 \times 18^2 +3 \times 18 +1 \le 18^3 \iff 3 \times 18 +1 \le 15 \times 18^2 ）$

then assume $abcd \not =0$,let $x_1=\dfrac{1}{a},x_2=\dfrac{1}{b},x_3=\dfrac{1}{c},x_4=\dfrac{1}{4} \to \sum_{cyc} x_{1}=4x_1x_2x_3x_4 \implies 19x_1x_2x_3x_4 \le 3\sum_{cyc}x_1x_2+1$

$f=19x_1x_2x_3x_4-3\sum_{cyc}x_1x_2-1,g=x_1+x_2+x_3+x_4-4x_1x_2x_3x_4 , F=f+\lambda g $

$F_{x_1}=19x_2x_3x_4-3(x_2+x_3+x_4)+\lambda (1-4x_2x_3x_4)=0$ $F_{x_2}=19x_1x_3x_4-3(x_1+x_3+x_4)+\lambda (1-4x_1x_3x_4)=0$ $F_{x_3}=19x_2x_1x_4-3(x_2+x_1+x_4)+\lambda (1-4x_2x_1x_4)=0$ $F_{x_4}=19x_2x_3x_1-3(x_2+x_3+x_1)+\lambda (1-4x_2x_3x_1)=0$

$(x_1-x_2)(\lambda-3(x_3+x_4))=0$

$(x_2-x_3)(\lambda-3(x_1+x_4))=0$

$(x_3-x_4)(\lambda-3(x_1+x_2))=0$

case I: $x_1=x_2=x_3=x_4=1$

case II: $x_1=x_2=x_3=x , \lambda = 6x_1=6x$

$19x^3-9x+6x-24x^4=0 \to 24x^3-19x^2+3=0 \to (3x+1)(8x^2-9x+3)=0 \to $ \ $x=-\dfrac{1}{3}$, ignore.

case III: $x_1=x_2, \lambda=3(x_1+x_4)=3(x_1+x_2) \implies$ case II, ignore

case IV: $\lambda=3(x_1+x_2)=3(x_2+x_3)=3(x_3+x_4) \to x_1=x_3=x,x_2=x_4=y $

$19xy-3-12x^2y-12xy^2=0,x+y=2x^2y^2 \to 19xy-3-24x^3y^3=0 \to xy<1 \implies x,y $ have no real root. ignore.

so only case one is OK.

$\dfrac{\partial f}{\partial x_1}=19x_2x_3x_4-3(x_2+x_3+x_4) ,\dfrac{\partial ^2 f}{\partial ^2 x_1}=0 , \dfrac{\partial ^2 f}{\partial x_1 \partial x_2}=19x_3x_4-3=\dfrac{\partial ^2 f}{\partial x_2 \partial x_1}=16$

$H=\left[\array{0 & 16 & 16 &16 \\ 16 & 0 & 16 &16 \\ 16 & 16 & 0 & 16 \\16 & 16 &16 & 0 }\right] <0 $

since there is only one solution, so it will be globe max also.

$f_{max}=19-3 \times 6 -1=0 $

QED.

chenbai
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Let $a+b+c+d=3u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.

Hence, we need to prove that $$\prod_{cyc}(a+3)\geq\sum_{cyc}(a+3)(b+3)(c+3)$$ or $$3(ab+ac+bc+ad+bd+cd)+2(abc+abd+acd+bcd)+abcd\geq27$$ or $$18v^2w^2+8w^4+t^4\geq27w^4$$ or $$19w^4-18v^2w^2-t^4\leq0$$ or $$w^2\leq\frac{9v^2+\sqrt{81v^4+19t^4}}{19}.$$ But $$24(3v^4-4uw^3+t^4)=(a-b)^2(c-d)^2+(a-c)^2(b-d)^2+(a-d)^2(b-c)^2\geq0,$$ which gives $t^4\geq4uw^3-3v^4$.

Hence, it remains to prove that $$w^2\leq\frac{9v^2+\sqrt{81v^4+19(4uw^3-3v^4)}}{19}$$ or $$9v^2+\sqrt{24v^6+76uw^3}\geq19w^2,$$ which is obvious by Maclaurin: $u\geq w$ and $v^2\geq w^2$.

A proof of Maclaurin see here: How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$

Done!

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Michael Rozenberg
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