An Easy Inequality

For all non-negative reals x,y,zx,y,z, prove that x2+y2+z2xyz+yzx+zxyx^2+y^2+z^2\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}

Note by Daniel Liu
5 years, 2 months ago

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While Muirhead works, another viable option is a generalization of rearrangement by taking four copies of {x,y,z}\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\}. (WLOG xyz0x\geq y\geq z\geq 0) Since {x,y,z}\{\sqrt{x},\,\sqrt{y},\,\sqrt{z}\} is monotonic decreasing, we have

xxxx+yyyy+zzzz\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x} + \sqrt{y}\cdot\sqrt{y}\cdot\sqrt{y}\cdot\sqrt{y} + \sqrt{z}\cdot\sqrt{z}\cdot\sqrt{z}\cdot\sqrt{z} xxyz+yyzx+zzxy\geq \sqrt{x}\cdot\sqrt{x}\cdot\sqrt{y}\cdot\sqrt{z} + \sqrt{y}\cdot\sqrt{y}\cdot\sqrt{z}\cdot\sqrt{x} + \sqrt{z}\cdot\sqrt{z}\cdot\sqrt{x}\cdot\sqrt{y}

Bob Krueger - 5 years, 2 months ago

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A solution by Muirhead that I discovered while writing the problem is as follows:

Note that (2,0,0)(1,12,12)(2,0,0)\succ \left(1,\dfrac{1}{2},\dfrac{1}{2}\right).

Thus, by Muirhead, 2(x2+y2+z2)2(xyz+yzx+zxy)2(x^2+y^2+z^2)\ge 2(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}) and the result follows.

The challenge I give you is to try to prove it using AM-GM. It shouldn't be too difficult.

Daniel Liu - 5 years, 2 months ago

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By AM-GM,

x2+x2+y2+z24x4y2z24\dfrac{x^2 + x^2 + y^2 + z^2}{4} \ge \sqrt[4]{x^4y^2z^2}

so 12x2+14y2+14z2xyz.\dfrac12x^2 + \dfrac14y^2 + \dfrac14z^2 \ge x\sqrt{yz}. Similarly, if we duplicate the y2y^2 and z2z^2 terms, we get

14x2+12y2+14z2yzx,\dfrac14x^2 + \dfrac12y^2 + \dfrac14z^2 \ge y\sqrt{zx}, and

14x2+14y2+12z2zxy.\dfrac14x^2 + \dfrac14y^2 + \dfrac12z^2 \ge z\sqrt{xy}. Adding these inequalities together, we have

x2+y2+z2xyz+yzx+zxyx^2+y^2+z^2 \ge x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy} as requested. \square

Michael Tang - 5 years, 2 months ago

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Ah, you beat me to it with the same solution I posted. Kudos to you on being first.

Trevor B. - 5 years, 2 months ago

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By AM-GM, 2x2+y2+z24=x2+x2+y2+z24x4y2z24.\dfrac{2x^2+y^2+z^2}{4}=\dfrac{x^2+x^2+y^2+z^2}{4}\ge\sqrt[4]{x^4y^2z^2}.

Similarly, x2+2y2+z24x2y4z24\dfrac{x^2+2y^2+z^2}{4}\ge\sqrt[4]{x^2y^4z^2} and x2+y2+2z24x2y2z44\dfrac{x^2+y^2+2z^2}{4}\ge\sqrt[4]{x^2y^2z^4}. Note that x4y2z24=x2yz=xyz,\sqrt[4]{x^4y^2z^2}=\sqrt{x^2yz}=x\sqrt{yz}, which can also apply to the other three radicals.

Dividing through, we have this system of inequalities.

12x2+14y2+14z2xyz14x2+12y2+14z2yzx14x2+14y2+12z2zxy \begin{aligned} \dfrac{1}{2}x^2+\dfrac{1}{4}y^2+\dfrac{1}{4}z^2&\ge x\sqrt{yz}\\ \dfrac{1}{4}x^2+\dfrac{1}{2}y^2+\dfrac{1}{4}z^2&\ge y\sqrt{zx}\\ \dfrac{1}{4}x^2+\dfrac{1}{4}y^2+\dfrac{1}{2}z^2&\ge z\sqrt{xy} \end{aligned}

Adding these together yields x2+y2+z2xyz+yzx+zxy,x^2+y^2+z^2\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}, as desired.

Q.E.D.\mathbb{Q.E.D.}


@Daniel Liu, Can you please explain your Muirhead solution a little further? I'm not familiar with that.

Trevor B. - 5 years, 2 months ago

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Muirhead states that for any two sets of numbers a1,a2,ana_1,a_2,\ldots a_n and b1,b2,bnb_1,b_2,\ldots b_n, if (a1,a2,an)(b1,b2,bn)(a_1,a_2,\ldots a_n)\succ(b_1,b_2,\ldots b_n), then symx1a1x2a2xnansymx1b1x2b2xnbn\sum_{sym}x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}\ge \sum_{sym}x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}

Search up Majorization if you do not know what \succ (which means majorizes) means.

But Muirhead is basically just a generalized AM-GM. Setting the first set of numbers as 1,0,01,0,\ldots 0 and the second set as 1n,1n,1n\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}, we see that (1,0,0)(1n,1n,1n)(1,0,\ldots 0)\succ \left(\dfrac{1}{n},\dfrac{1}{n},\ldots \dfrac{1}{n}\right) and using Muirhead's gives us the familiar nn-term AM-GM.

Daniel Liu - 5 years, 2 months ago

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We could apply AM-GM on yzy+z2\sqrt {yz}\le \frac {y+z}{2} which would lead to proving the well known x2+y2+z2xy+yz+xzx^2+y^2+z^2\ge xy+yz+xz

Xuming Liang - 5 years, 2 months ago

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It is also true that xy+yz+zxxyz+yzx+zxyxy+yz+zx\ge x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}

Can you prove that?

Daniel Liu - 5 years, 2 months ago

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Well by doing what I said this is actually what you get :).....: xyz+yzx+zxyxy+z2+yx+z2+zy+x2=xy+yz+zxx\sqrt {yz}+y\sqrt {zx}+z\sqrt {xy}\le x\frac {y+z}{2}+ y\frac {x+z}{2}+z\frac {y+x}{2}=xy+yz+zx

Xuming Liang - 5 years, 2 months ago

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@Xuming Liang Ah, I misunderstood your statement. Sorry.

However, it is also true that (xyz+yzx+zxy)327x2y2z2(x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy})^3\ge 27x^2y^2z^2

How about that one?

Daniel Liu - 5 years, 2 months ago

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@Daniel Liu No need to apologize :) Anyway this one is just immediate AM-GM

Xuming Liang - 5 years, 2 months ago

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@Xuming Liang It's good to finally see some inequality actions going on here, I've never been interested in inequalities until recently :) Here's one I feel kinda proud for solving it:

a,b,ca,b,c are positive reals, ab+bc+ca=1ab+bc+ca=1 prove 1a+6b3+1b+6c3+1c+6a31abc. \sqrt[3]{\frac{1}{a}+6b}+\sqrt[3]{\frac{1}{b}+6c}+\sqrt[3]{\frac{1}{c}+6a }\leq\frac{1}{abc}.

P.S. Good think on AoPS there's a quick way for you to copy and paste latex..

Xuming Liang - 5 years, 2 months ago

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@Xuming Liang lol yea no need to do any clever separating of terms.

How about this one?

2xyz+2yzx+2zxysymxy2z332x\sqrt{yz}+2y\sqrt{zx}+2z\sqrt{xy}\ge \sum_{sym}\sqrt[3]{xy^2z^3}

Daniel Liu - 5 years, 2 months ago

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@Michael Tang @Trevor B. Try your hand at this one.

Daniel Liu - 5 years, 2 months ago

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@Daniel Liu I'm neither Michael ,nor Trevor, but since

1,12,121,\frac{1}{2},\frac{1}{2} majorizes 1,23,131,\frac{2}{3},\frac{1}{3}, use Muirhead to obtain the result.

Bogdan Simeonov - 5 years, 2 months ago

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@Bogdan Simeonov Sorry, but the point was to prove it using AM-GM @Bogdan Simeonov

Since anything provable using Muirhead (which was the instakill solution that I suggested) could be proven using AM-GM, the challenge was to prove my suggested inequalities using AM-GM.

Daniel Liu - 5 years, 2 months ago

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@Daniel Liu Oh, ok :D

Bogdan Simeonov - 5 years, 2 months ago

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