IIT JEE , AIEEE Forum - Mathematics , Algebra - For the Inequality Enthusiasts

hsbhatt

If x,y and z are non-negative reals prove that:

$\frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}$

akki_0410

As, x,y,z are real positive no's.

solving RHS,

multiplying all 3 equn, and taking its square root, we get,

solving LHL,

.......(ii)

similarly,

........(iii)

taking AM greater than equal to GM of (ii) and (iii), we get,

as, LHL & RHL ar both greater than root (xyz)

so we cannot comment on sign between LHL and RHL.................

PLEASE CHECK WEATHER I AM CORRECT OR NOT...................

rudra.panda

first we don't have to use AM-GM inequality or the sign will be other way around.

hsbhatt

@akki: You are absolutely right that nothing can be concluded by this approach Very Happy .

But do keep trying. Its very instructive

ronty

firstly the power-mean inequality............

L.H.S.<={((x.y^1/2+y.z^1/2+xz^1/2)²+(yx^1/2+zy^1/2+zx^1/2)²)/18}^1/2

now we expand the squares...

L.H.S.<=square root of the following expression:

1/8{x²y+y²x+x²z+z²x+y²z+z²y+2xyz} -

1/18{5/4(x²y+y²x+x²z+z²x+y²z+z²y+2xyz) - {2x²(yz)^1/2+2xy(yz)^1/2+2yz(xy)^1/2+2z²(xy)^1/2+2xyz}}.

now,square root of 1/8{x²y+y²x+x²z+z²x+y²z+z²y+2xyz} is the reqd. R.H.S.

so all we need to do is to show that

X={5/4(x²y+y²x+x²z+z²x+y²z+z²y+2xyz) - {2x²(yz)^1/2+2xy(yz)^1/2+2yz(xy)^1/2+2z²(xy)^1/2+2xyz}>=0.

since 2(yz)^1/2<=y+z,and similarly the others , we get

X>=5/4(x²y+y²x+x²z+z²x+y²z+z²y+2xyz) -(x²(y+z)+xy(y+z)+yz(x+y)+z²(x+y) -xyz/2)

=1/4(x²y+y²x+x²z+z²x+y²z+z²y) -3xyz/2.

by simple A.M. G.M. inequality, we see that the last expression>=0.

hence we are done.

sandeepramesh

2 AM-GM's will do it.....

hsbhatt

nice work ronty. But could I trouble you to latex your solution. It will be easy for others to read it.

hsbhatt

It is very good to see you guys doing these problems. Because frankly, I have only recently begun to learn the subject of inequalities and definitely did not know beyond AM-GM when I attended JEE. This problem was given as part of a tech-fest called Shaastra that just concluded at IIT-Chennai. I had the following solution:

We have by Cauchy Schwarz Inequality:

$\sqrt{(xy+yz+zx)(x+y+z)} \ge x \sqrt y + y \sqrt z + z \sqrt x$

Again by the application of the same inequality,

$\sqrt{(xy+yz+zx)(y+z+x)} \ge y \sqrtx + z \sqrt y + x \sqrt z$

Hence, we get

$\sqrt{(xy+yz+zx)(x+y+z)} \ge \dfrac{x \sqrt y + y \sqrt z + z \sqrt x}{2} + \dfrac{y \sqrtx + z \sqrt y + x \sqrt z}{2}$

Now, we only have to prove that

$\frac{1}{3}\sqrt{(xy+yz+zx)(x+y+z)} \le \sqrt{\left(\frac{x+y}{2} \right) \left(\frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}$

Or, squaring both sides,

$8(xy+yz+zx)(x+y+z)} \le 9 (x+y)(y+z)(z+x)$

We can simplify things by exploiting homogeneity and hence letting x+y+z = 1

So that we now have to prove that

$8(xy+yz+zx)(x+y+z)} \le 9 (1-x)(1-y)(1-z) \\ \\<br/>= 9(1 - (x+y+z)+ \sum xy -xyz) = 9(xy+y+z+zx-xyz)$

Or

$xy+yz+zx \ge 9 \ \text{or} \ \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \ge 9$

But this is obviously true as we have from AM-GM

$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \ge 9$

And since x+y+z = 1, we do have

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \ge 9$

Hence it follows that

$\frac{\frac{x \sqrt y + y \sqrt z + z \sqrt x}{3} + \frac{y \sqrt x + z \sqrt y + x \sqrt z}{3}}{2} \le \sqrt{\left(\frac{x+y}{2}\right) \left( \frac{y+z}{2} \right) \left(\frac{z+x}{2}\right)}$

jishnudas1991

jishnudas1991

Re:For the Inequality Enthusiasts - I

hsbhatt

The last part of Jishnu's post answers an extension of this question and was asked in the techfest from where I filched the question.

Good work guys.

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