Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Question

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$ ?

I tried something basic like $(x+y+z)^2 = x^2+y^2+z^2 + 2xy+2zx+2yz$, so we just need to prove $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} + 2xy+2zx+2yz >= 9$, but nothing there...

I was also thinking of Lagrange multiplier (and I know it's generally not used in contest math), is there an easy way to prove the only solution to $\nabla f(x,y,z) = \lambda \nabla g(x,y,z)$ is $x=y=z$? obviously here $f(x,y,z) = \frac{x}{y}+\frac{y}{z}+\frac{z}{x} - (x^2+y^2+z^2)$, $g(x,y,z)=(x+y+z-3)$

Answer 1

This "solution" is rather a story about searching for the solution in this case and in similar cases. It follows standard ideas to solve such inequalities.

(I decided to wait till the one or the other one answer of Michael Rozenberg was accepted, then post the thoughts towards a solution, based on a paperless discussion, where i could not show the details.)

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First, one may take a look maybe at

Kedlaya, Inequalities

to see usual techniques applied in similar cases.

The reader short in time may please go straightforward to the proof, marked almost at the end of the post. (It is a version of one of the proofs of Michael Rozenberg. )

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First we obtain a homogenized version, then we clear denominators, and in the polynomial inequality obtained we use domination of a term $x^ay^bz^c$, figured as the point $(a,b,c)$ in the space, by seing this point inside some convex figure with extremal points $(a',b',c')$ that can be used to dominate. Here are the steps explicitly, no artificial recipes.

(1) We show the homogenized version: $$ \frac xy+\frac yz+\frac zx\ge \frac{9(x^2+y^2+z^2)}{(x+y+z)^2}\ . $$ (2) After eliminating denominators, we have to show equivalently: $$ (x^2z+y^ 2x+z^2y)(x+y+z)^2 - 9xyz(x^2+y^2+z^2)\ge 0\ . $$ We eliminate all parentheses, have to show that $$ \sum(x^4z+x^3y^2+2x^3z^2+3x^2y^2z)\ge \sum 7x^3yz\ , $$ i.e. $$ \underbrace{ \sum\Big(\ x^4z+x^3y^2+2x^3z^2+3x^2y^2z-7x^3yz\ \Big) }_{(*)}\ge 0\ , $$

(3) We make a picture of the points, and mark the weightings of them. To have a picture in $2D$, rather than in $3D$, we will restrict to the plane $a+b+c=5$, and $(a,b,c)$ corresponds to $x^ay^bz^c$.

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The points corresponding to (minus seven times) $x^3yz$ (and cyclic relatives) have to be "covered" with a linear combination of the others with coefficients $\ge 0$. (We would like to reduce the problem to a problem in convex analysis, but things are not so easy in the given situation, because...)

"Bunching" can not be applied now, since the terms in (three times) $\sum x^2y^2z$ are "weaker" than the terms in (minus seven times) $\sum x^3yz$.

So a tentative solution may like to engage the terms like $x^2y^2z$ (and cyclic relatives) and those in $-x^3yz$ in an inequality which steals the pattern of the Schur inequality. We have to search for / to build squares with these "weaker" terms, and in the same time (use also the otehrs) to control the negative terms.

So we are searching for an expression of the shape $$ \sum_{\text{cyclic}} xP(x,y,z)^2\ , $$ which is "as close as possible" to our expression $(*)$, which - after possible cancellations - still contains the terms $+x^2y^2z$ and terms $-x^3yz$. (Further positive terms are welcome.)

This is a good moment to stop reading here, switch to the proof of Michael Rozenberg, that realizes this task, appreciate and give credit for it.

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If not, then here is what i would do in such a situation. We must use each term with positive coefficients to control the three terms with negative coefficients. When building squares of monomials $A,B$ of the shape $(A-B)^2=A^2-2AB+B^2$, the negative term $2AB$ has multi-degree which is the "mid point" of the degrees of $A^2$ and $B^2$. And we indeed search for such possibilities to depict the negative terms as mid points of segments. And we do this only in the grey part where $x$ is a factor, since we need the $P^2$ in the term $\color{red}{x}P(x,y,z)^2$ in a picture:

$(x)$">

Some segments with midpoints in the nodes marked with $x^3yxZ$, $xy^3z$, $xyz^3$ were searched in the gray part and marked. Only the red ones involve also the must-be used terms $x^2y^2z$ (+ cycled). So we build $$ \begin{aligned} &E(x,y,z) \\ &\qquad:= \sum\Big(\ (x^4z+x^3y^2+2x^3z^2+3x^2y^2z - 7x^3yz) - xy^2(y-z\sqrt 3)^2\ \Big) \\ &\qquad:= \sum\Big(\ x^3y^2+2x^3z^2 - (7-2\sqrt 3)x^3yz\ \Big)\ , \end{aligned} $$ and "unfortunately" we do not have enough power to dominate with $x^3y^2+2x^3z^2$ more than $2\sqrt 2 x^3yz$. It is important to know in this case, and in similar cases why we fail, a good reason. Well, the given inequality becomes equality for $x=y=z$, but in our attempt we use $(y-z\sqrt 3z)^2$ which is not optimal near $x=y=z$. (But something like $(x-y)^2$, or $(y-z)^2, or $(z-x)^2$, or $(2x-y-z)^2$, or... would be.)

We go back and see the point we were wasting material, it is the place where we use $x\color{red}{y^2}$ as a factor, and this is a "big" deviation.

After making once for all times the above experience, the solver goes directly to...

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So we have to search for slightly more complicated squares, we look at the term $xy^4$, and try to find rewrite our expression $(*)$ in the shape $$\sum x(y-ax)^2(y-bz)^2\ .$$

(We cannot also add a $z$-term inside $(y-ax)^2$, since this would generate some $xz^4$ in the final, we do not have this term, and it would be hard to control. Same holds for the second parenthesis, inside $(y-bz)^2$ we cannot add any $x$-term, else we introduce $x^5$ after expanding.)

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In such occasions i am using computer aid, here sage.

sage: var('x,y,z,a,b');
sage: E = ( ( x/y + y/z + z/x ) - 9*(x^2+y^2+z^2)/(x+y+z)^2 ).factor().numerator()
sage: E
x^3*y^2 + 2*x^2*y^3 + x*y^4 + x^4*z - 7*x^3*y*z + 3*x^2*y^2*z - 7*x*y^3*z + 2*x^3*z^2 + 3*x^2*y*z^2 + 3*x*y^2*z^2 + y^3*z^2 + x^2*z^3 - 7*x*y*z^3 + 2*y^2*z^3 + y*z^4
sage: v(x,y,z) = x * (y-a*x)^2 * (y-b*z)^2
sage: F = ( E - v(x,y,z) - v(y,z,x) - v(z,x,y) ).expand()
sage: F.coefficient( x^3*y^2 )
-a^2 + 1

So it may be a good idea to use $a=\pm1$. We set then $a=1$, and look for the coefficient that makes problems, the one in $x^3yz$:

sage: F1 = F.subs( { a : 1, b : 7/4 } )
sage: F1 = F.subs( { a : 1} )
sage: F1.coefficient( x^3*y*z )
4*b - 7

So it is a good idea to take $b=7/4$, and look at the remained terms.

sage: F.subs( { a : 1, b : 7/4 } )
15/16*x^2*y^3 - 15/16*x^2*y^2*z + 15/16*x^3*z^2 - 15/16*x^2*y*z^2 - 15/16*x*y^2*z^2 + 15/16*y^2*z^3

This can be controlled by bunching. But because i hate fractions, and because it (also) works, we take $b=2$, and obtain the following proof:

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Proof of the inequality in the OP:

$$ \begin{aligned} & \operatorname{Numerator of}\left[\ \left(\frac xy+\frac yz+\frac zx\right) -\frac {9(x^2+y^2+z^2)}{(x+y+z)^2}\ \right] \\ &\qquad = \sum_{\text{cyclic}} x(y-x)^2(y-2z)^2 + \sum_{\text{cyclic}} (x^3yz-x^2y^2z) \\ &\qquad = \sum_{\text{cyclic}} x(y-x)^2(y-2z)^2 + \frac 12xyz\sum_{\text{cyclic}}(x-y)^2 \\ &\qquad\ge 0 \ . \end{aligned} $$

$\square$

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Computer check of the used polynomial equality:

sage: var('x,y,z');
sage: E = ( ( x/y + y/z + z/x ) - 9*(x^2+y^2+z^2)/(x+y+z)^2 ).factor().numerator()
sage: h(x, y, z) = x *(y-x)^2 * (y-2*z)^2   +   x*y*z * (x-y)^2 / 2
sage: ( E - h(x,y,z) - h(y,z,x) - h(z,x,y) ).expand()
0

Answer 2

Let $x+y+z=3u$, $xy+xz+yz=3v^2$, $xyz=w^3$ and $u^2=tv^2$.

Hence, $t\geq1$ and we need to prove that $$u^2\sum_{cyc}x^2z\geq3(3u^2-2v^2)w^3$$ or $$u^2\sum_{cyc}2x^2z\geq6(3u^2-2v^2)w^3$$ or $$u^2\sum_{cyc}(x^2y+x^2z)-6(3u^2-2v^2)w^3\geq u^2\sum_{cyc}(x^2y-x^2z)$$ or $$u^2(9uv^2-3w^3)-6(3u^2-2v^2)w^3\geq u^2(x-y)(x-z)(y-z)$$ or $$3(3u^3v^2-7u^2w^3+4v^2w^3)\geq u^2(x-y)(x-z)(y-z).$$ We'll prove that $3u^3v^2-7u^2w^3+4v^2w^3\geq0$.

Indeed, it's $f(w^3)\geq0,$ where $f$ increases, which says that it's enough to prove the last inequality for a maximal value of $w^3$, which happens for equality case of two variables.

Since the last inequality is homogeneous, it's enough to assume $y=z=1,$ which gives

$$\frac{(x+2)^3(2x+1)}{27}-\frac{7(x+2)^2x}{9}+\frac{4(2x+1)x}{3}\geq0$$ or $$(x-1)^2(x^2-2x+4)\geq0,$$ which is obvious.

Hence, it remains to prove that $$9(3u^3v^2-7u^2w^3+4v^2w^3)^2\geq u^4(x-y)^2(x-z)^2(y-z)^2$$ or $$(3u^3v^2-7u^2w^3+4v^2w^3)^2\geq3u^4(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$(13u^4-14u^2v^2+4v^4)w^6+3(u^4-5u^2v^2+2v^4)u^3w^3+3u^4v^6\geq0.$$ Now, if $u^4-5u^2v^2+2v^4\geq0$ so the last inequality is obviously true.

Hence, it remains to prove our inequality for $$u^4-5u^2v^2+2v^4<0$$ or $$1\leq t<\frac{5+\sqrt{17}}{2}.$$ Id est, it remains to prove that $$3u^6(u^4-5u^2v^2+2v^4)^2-4(13u^4-14u^2v^2+4v^4)u^4v^6\leq0$$ or $$3t(t^2-5t+2)^2-4(13t^2-14t+4)\leq0$$ or $$(t-1)^2(3t^3-24t^2+36t-16)\leq0,$$ which is true for $1\leq t<\frac{5+\sqrt{17}}{2}$.

Done!

By the same way we can get a best estimation.

Answer 3

There is also the following way: $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-x^2-y^2-z^2=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{9(x^2+y^2+z^2)}{(x+y+z)^2}=$$ $$=\frac{\sum\limits_{cyc}x(x-y)^2(2y^2-7yz+8z^2)}{2xyz(x+y+z)^2}\geq0.$$ We can get it by the following play.

Let $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{9(x^2+y^2+z^2)}{(x+y+z)^2}=\frac{\sum\limits_{cyc}x(x-y)^2(y^2+ayz+bz^2)}{xyz(x+y+z)^2}.$$ Now, after full expanding we need to solve some system and it's impotent to check that $a^2-4b\leq0.$

Answer 4

Also, BW helps:

Let $x=\min\{x,y,z\},$ $y=x+u$ and $z=y+v$.

Thus, $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq\frac{9(x^2+y^2+z^2)}{(x+y+z)^2}$$ it's $$3(u^2-uv+v^2)x^3+9uv^2x^2+(u^4-5u^3v+12u^2v^2+uv^3+v^4)x+uv^2(u+v)^2\geq0,$$ which is obvious.