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let $x,y,z>0$ and such $xyz=1$ show that $$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$

My try: I will find a value of the $k$ such $$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$ note $\ln{x}+\ln{y}+\ln{z}=0$,so $$\sum_{cyc}\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{3}+k(\ln{x}+\ln{y}+\ln{z})=\dfrac{1}{3}$$ so let $$f(x)=\dfrac{1}{2(x+1)^2+1}-k\ln{x}-\dfrac{1}{9}$$ $$\Longrightarrow f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}-\dfrac{k}{x}$$ let $f'(1)=0\Longrightarrow k=-\dfrac{8}{81}$

so $$f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}+\dfrac{8}{81x}=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}$$ so note when $1>x>\dfrac{1}{2}$ then $$f'(x)=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}<0$$ $x>1,f'(x)>0$ so $$f(x)\ge f(1)=0$$ so if $x,y,z>\dfrac{1}{2}$ we have prove done.

But for other case,How prove it? Thank you

  • If x,y,z >0 and x,y,z<1, Then you can use trignometric transformation such as x = sinu ; y = secu ; z = cotu making xyz = 1. I computed for u = pi/4 and u = pi/3. The LHS are 0.3364 and .34552 both of them are greater than (1/3). Confirm the ranges for x,y,z? – Satish Ramanathan Nov 07 '13 at 18:05
  • @satishramanathan But LHS is supposed to be greater by the inequality. Also note your substitution does not cover all cases, for e.g. what if I want $x = 1, y = \sec t, z = \cos t$ - for some $t$ - do you think you can find me a suitable $u$? – Macavity Nov 07 '13 at 19:22
  • @Macavity, I thought about even before you asked, x=1/20, y = 4, z = 5, then I can't find a u that will satisfy these three values of x,y and z. You are right, it cannot cover any value of x,y and z. The opposite is true, for any value of u, I can find an x,y and z satisfying the inequality and the condition that xyz=1. By the bye I have edited RHS to LHS. I will wait to appreciate anyone posting a solution that covers all x,y,z. Thanks for pointing it out – Satish Ramanathan Nov 07 '13 at 20:16

3 Answers3

2

$z=\dfrac{1}{xy}$, put in LHS and and clean the denominators, we have:

edit:

LHS-RHS=$ 9y^2x^4-8y^3x^3+2y^2x^3+9y^4x^2+2y^3x^2-9y^2x^2-8yx^2-8y^2x+2yx+9 \ge0 \iff $ $4y^2x^4-8yx^2+4\ge 0,\\4y^4x^2-8y^2x+4\ge0,\\5y^2x^4+5y^4x^2\ge 10x^3y^3,\\2y^2x^3+2y^3x^2\ge 4(xy)^{\frac{5}{2}} \iff\\ LHS \ge 2x^3y^3+4(xy)^{\frac{5}{2}}-9x^2y^2+2xy+1=2t^6+4t^5-9t^4+2t^2+1=(t-1)^2(2t^4+8t^3+5t^2+2t+1) \ge0, t=\sqrt{xy}$

the "=" will hold when $xy=1,x=y,y^2x=yx^2=1 \implies x=y=z=1$

chenbai
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  • Does it mean that for this inequality to be true, do x,y,z have to be 1? It is not as clean as one would like for any value of x,y,z. I am expecting a solution that would hold this inequality true for any x,y,z. Let me know. Thanks – Satish Ramanathan Nov 17 '13 at 16:12
  • @satishramanathan, this proof show the inequality is true as each step can be deducted reversely.I have a minor typo and I corrected it. – chenbai Nov 18 '13 at 03:19
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let $$x=\dfrac{bc}{a^2},y=\dfrac{ca}{b^2},z=\dfrac{ab}{c^2}$$ then we only prove follow inequality $$\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}+\dfrac{b^4}{3b^4+2c^2a^2+4b^2ca}+\dfrac{c^4}{4c^4+2a^2b^2+4c^2ab}\ge\dfrac{1}{3}$$ By Cauchy-Schwarz inequality \begin{align*} \sum_{cyc}\dfrac{a^4}{3a^4+2b^2c^2+4a^2bc}&\ge\dfrac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2)+4abc(a+b+c)}\\ &\ge\dfrac{(a^2+b^2+c^2)^2}{3(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)+4(a^2b^2+b^2c^2+c^2a^2)}\\ &=\dfrac{(a^2+b^2+c^2)^2}{3(a^2+b^2+c^2)^2}=\dfrac{1}{3} \end{align*}

math110
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-1

Hint:

Use Tringnometric transformation such that

x=sinu, y=secu and z = cotu

For any value of u you can find x,y,z >0 such that xyz = 1 by definition.

For a value of u = pi/6, x=.5, y = 1.154 and z = 1.732

Further, I computed LHS, for u = pi/4 and its value is .3365

Just to double check, I computed LHS for u = pi/3 and its value is .34552

In both sample cases, LHS is >= (1/3).

Thanks

Satish

Satish Ramanathan
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