If $x,y,z>0.$Prove: $$(x+y+z) \left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\geq9\sqrt[]\frac{x^2+y^2+z^2}{xy+yz+zx}$$ I was not able to solve this problem instead I could solve similar inequality when we have two variable.I assumed $y=tx$ and uesd derivative. Can this be generalized as:
If ${a_i>0}\quad(i=1,2,...,n)$ $$\sum_{i=1}^n a_{i} \sum_{i=1}^n \frac{1}{a_{i}}\geq n^2\sqrt[]\frac{\sum_{i=1}^n a^2_{i} }{\sum_{i=1}^n a_{i}a_{i+1} }$$ $a_{n+1}=a_{1}$
Question from Jalil Hajimir
On
Final result: We know that this inequality holds for all $n\le 4$. For $n=5$, @RiverLi has provided a counter-example: $a_1=a_3=1,a_2=8,a_4=a_5=2$. For this we have LHS-RHS$\ \approx-0.148$.
For $n=2$: We have $$(x+y)^2\left(\frac1x+\frac1y\right)^2-\frac{4^2(x^2+y^2)}{xy+yx}=\frac{(x - y)^4}{x^2 y^2}\geq0.$$
For $n=3:$
We need to prove $$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2 (x+y+z)^2-\frac{9^2 \left(x^2+y^2+z^2\right)}{x y+x z+y z}\geq0.$$
After a full expansion we see that this is the same as $$\frac1{x^2 y^2 z^2 (x (y+z)+y z)}\cdot X\geq0,$$
where $$X={x^5 (y+z)^3+x^4 \left(2 y^4+11 y^3 z-63 y^2 z^2+11 y z^3+2 z^4\right)+x^3 \left(y^5+11 y^4 z+31 y^3 z^2+31 y^2 z^3+11 y z^4+z^5\right)+x^2 y z \left(3 y^4-63 y^3 z+31 y^2 z^2-63 y z^3+3 z^4\right)+x y^2 z^2 \left(3 y^3+11 y^2 z+11 y z^2+3 z^3\right)+y^3 z^3 (y+z)^2}.$$
So we only have to prove $X\geq 0$. By symmetry of the inequality [Footnote 1], we can assume that $x\le y\le z$. So there exist $a,b\geq 0$ such that $y=x+a$ and $z=x+a+b$. Using this we see $$X=4 a^8+16 a^7 b+60 a^7 x+25 a^6 b^2+210 a^6 b x+219 a^6 x^2+19 a^5 b^3+282 a^5 b^2 x+657 a^5 b x^2+364 a^5 x^3+7 a^4 b^4+180 a^4 b^3 x+717 a^4 b^2 x^2+910 a^4 b x^3+318 a^4 x^4+a^3 b^5+54 a^3 b^4 x+339 a^3 b^3 x^2+748 a^3 b^2 x^3+636 a^3 b x^4+144 a^3 x^5+6 a^2 b^5 x+72 a^2 b^4 x^2+212 a^2 b^3 x^3+369 a^2 b^2 x^4+216 a^2 b x^5+27 a^2 x^6+12 a b^5 x^2+26 a b^4 x^3+51 a b^3 x^4+108 a b^2 x^5+27 a b x^6+8 b^5 x^3+3 b^4 x^4+18 b^3 x^5+27 b^2 x^6$$
and hence $X\geq 0$ because of $x,a,b\geq 0$. It follows that the original inequality is correct.
For $n=4$: Here is a computer-assisted proof for $4$ variables $w,x,y,z$ (sadly, this proof is very hard to come up with - and even very hard to check without the help of a computer):
By the same method as above we get that the inequality is equivalent to $$X\geq 0,$$ where
$$X=\text{see below}$$
Note that the only negative terms are the last three terms. Let $Y$ be the expression that we obtain by setting $b=0$ in $X$. Then $X\geq Y$ and $$Y=(a + w)^2 (18 a^8 + 3 a^7 (19 c + 94 w) + 2 a^6 (34 c^2 + 398 c w + 529 w^2) + 2 c^2 w^3 (9 c^3 - 14 c^2 w + 32 c w^2 + 64 w^3) + a c^2 w^2 (21 c^3 - 58 c^2 w + 16 c w^2 + 320 w^3) + a^5 (38 c^3 + 840 c^2 w + 2461 c w^2 + 1790 w^3) + 2 a^4 (5 c^4 + 206 c^3 w + 1028 c^2 w^2 + 1605 c w^3 + 786 w^4) + 2 a^2 w (4 c^5 + 31 c^4 w + 130 c^3 w^2 + 428 c^2 w^3 + 224 c w^4 + 64 w^5) + a^3 (c^5 + 94 c^4 w + 678 c^3 w^2 + 1948 c^2 w^3 + 1936 c w^4 + 704 w^5)).$$
So it remains to show that
Proof of 1. Simply use $c^3+c w^2\geq 2c^2 w$ by AM-GM. $\square$
Proof of 2. We have $$21 c^3+320w^3=\frac{21}2 c^3+\frac{21}2 c^3+320 w^3\overset{\text{AM-GM}}\geq 3c^2 w\sqrt[3]{320\frac{21^2}4}\geq 98.3c^2 w. \square$$
Here is the expression for $X$ when $n=4$: $$X=18 a^{10}+123 b a^9+57 c a^9+318 w a^9+365 b^2 a^8+68 c^2 a^8+1640 w^2 a^8+341 b c a^8+1952 b w a^8+910 c w a^8+614 b^3 a^7+38 c^3 a^7+4188 w^3 a^7+352 b c^2 a^7+9010 b w^2 a^7+4110 c w^2 a^7+868 b^2 c a^7+5144 b^2 w a^7+976 c^2 w a^7+4836 b c w a^7+640 b^4 a^6+10 c^4 a^6+6210 w^4 a^6+168 b c^3 a^6+20388 b w^3 a^6+8928 c w^3 a^6+754 b^2 c^2 a^6+20954 b^2 w^2 a^6+3804 c^2 w^2 a^6+19374 b c w^2 a^6+1218 b^3 c a^6+7576 b^3 w a^6+488 c^3 w a^6+4424 b c^2 w a^6+10780 b^2 c w a^6+423 b^5 a^5+c^5 a^5+5638 w^5 a^5+37 b c^4 a^5+26443 b w^4 a^5+10817 c w^4 a^5+295 b^2 c^3 a^5+41280 b^2 w^3 a^5+6900 c^2 w^3 a^5+37060 b c w^3 a^5+855 b^3 c^2 a^5+26814 b^3 w^2 a^5+1540 c^3 w^2 a^5+15038 b c^2 w^2 a^5+37720 b^2 c w^2 a^5+1017 b^4 c a^5+6790 b^4 w a^5+114 c^4 w a^5+1852 b c^3 w a^5+8146 b^2 c^2 w a^5+13008 b^3 c w a^5+173 b^6 a^4+3108 w^6 a^4+3 b c^5 a^4+20660 b w^5 a^4+7530 c w^5 a^4+51 b^2 c^4 a^4+45873 b^2 w^4 a^4+6808 c^2 w^4 a^4+39461 b c w^4 a^4+257 b^3 c^3 a^4+45132 b^3 w^3 a^4+2028 c^3 w^3 a^4+23532 b c^2 w^3 a^4+62348 b^2 c w^3 a^4+541 b^4 c^2 a^4+20570 b^4 w^2 a^4+260 c^4 w^2 a^4+5030 b c^3 w^2 a^4+23898 b^2 c^2 w^2 a^4+39170 b^3 c w^2 a^4+505 b^5 c a^4+3768 b^5 w a^4+10 c^5 w a^4+352 b c^4 w a^4+2722 b^2 c^3 w a^4+7748 b^3 c^2 w a^4+9118 b^4 c w a^4+40 b^7 a^3+960 w^7 a^3+9600 b w^6 a^3+2832 c w^6 a^3+3 b^2 c^5 a^3+30240 b^2 w^5 a^3+3980 c^2 w^5 a^3+24640 b c w^5 a^3+31 b^3 c^4 a^3+42048 b^3 w^4 a^3+1214 c^3 w^4 a^3+19934 b c^2 w^4 a^3+56972 b^2 c w^4 a^3+111 b^4 c^3 a^3+29068 b^4 w^3 a^3+160 c^4 w^3 a^3+5724 b c^3 w^3 a^3+32056 b^2 c^2 w^3 a^3+54964 b^3 c w^3 a^3+181 b^5 c^2 a^3+9672 b^5 w^2 a^3+38 c^5 w^2 a^3+760 b c^4 w^2 a^3+6478 b^2 c^3 w^2 a^3+19534 b^3 c^2 w^2 a^3+23390 b^4 c w^2 a^3+138 b^6 c a^3+1252 b^6 w a^3+24 b c^5 w a^3+390 b^2 c^4 w a^3+1916 b^3 c^3 w a^3+3974 b^4 c^2 w a^3+3676 b^5 c w a^3+4 b^8 a^2+128 w^8 a^2+2432 b w^7 a^2+448 c w^7 a^2+11776 b^2 w^6 a^2+1624 c^2 w^6 a^2+9152 b c w^6 a^2+b^3 c^5 a^2+22784 b^3 w^5 a^2+356 c^3 w^5 a^2+10056 b c^2 w^5 a^2+30624 b^2 c w^5 a^2+7 b^4 c^4 a^2+22104 b^4 w^4 a^2+3114 b c^3 w^4 a^2+23134 b^2 c^2 w^4 a^2+41792 b^3 c w^4 a^2+19 b^5 c^3 a^2+11272 b^5 w^3 a^2+68 c^5 w^3 a^2+640 b c^4 w^3 a^2+6704 b^2 c^3 w^3 a^2+22276 b^3 c^2 w^3 a^2+27324 b^4 c w^3 a^2+25 b^6 c^2 a^2+2716 b^6 w^2 a^2+68 b c^5 w^2 a^2+804 b^2 c^4 w^2 a^2+4066 b^3 c^3 w^2 a^2+8618 b^4 c^2 w^2 a^2+8004 b^5 c w^2 a^2+16 b^7 c a^2+224 b^7 w a^2+18 b^2 c^5 w a^2+180 b^3 c^4 w a^2+634 b^4 c^3 w a^2+1024 b^5 c^2 w a^2+776 b^6 c w a^2+256 b w^8 a+2560 b^2 w^7 a+576 c^2 w^7 a+2048 b c w^7 a+7040 b^3 w^6 a+144 c^3 w^6 a+3328 b c^2 w^6 a+9728 b^2 c w^6 a+9184 b^4 w^5 a+928 b c^3 w^5 a+9504 b^2 c^2 w^5 a+17664 b^3 c w^5 a+6544 b^5 w^4 a+57 c^5 w^4 a+231 b c^4 w^4 a+3452 b^2 c^3 w^4 a+12832 b^3 c^2 w^4 a+16032 b^4 c w^4 a+2480 b^6 w^3 a+80 b c^5 w^3 a+696 b^2 c^4 w^3 a+3644 b^3 c^3 w^3 a+7908 b^4 c^2 w^3 a+7360 b^5 c w^3 a+400 b^7 w^2 a+34 b^2 c^5 w^2 a+332 b^3 c^4 w^2 a+1154 b^4 c^3 w^2 a+1848 b^5 c^2 w^2 a+1392 b^6 c w^2 a+16 b^8 w a+4 b^3 c^5 w a+28 b^4 c^4 w a+76 b^5 c^3 w a+100 b^6 c^2 w a+64 b^7 c w a+256 b^2 w^8+128 c^2 w^8+256 b c w^8+1024 b^3 w^7+64 c^3 w^7+640 b c^2 w^7+1536 b^2 c w^7+1728 b^4 w^6+192 b c^3 w^6+1920 b^2 c^2 w^6+3456 b^3 c w^6+1600 b^5 w^5+18 c^5 w^5+36 b c^4 w^5+800 b^2 c^3 w^5+3200 b^3 c^2 w^5+4000 b^4 c w^5+848 b^6 w^4+33 b c^5 w^4+233 b^2 c^4 w^4+1248 b^3 c^3 w^4+2744 b^4 c^2 w^4+2544 b^5 c w^4+224 b^7 w^3+20 b^2 c^5 w^3+192 b^3 c^4 w^3+660 b^4 c^3 w^3+1048 b^5 c^2 w^3+784 b^6 c w^3+16 b^8 w^2+4 b^3 c^5 w^2+28 b^4 c^4 w^2+76 b^5 c^3 w^2+100 b^6 c^2 w^2+64 b^7 c w^2-(82 c^4 w^4 a^2+28 c^4 w^6+114 c^4 w^5 a)$$
[Footnote 1] For $x,y,z\in]0,\infty[$, let $P(x,y,z)$ denote the statement that "the inequality is true for the chosen $x,y,z$." It is left as an exercise to the reader to check that, whenever $\{a,b,c\}=\{x,y,z\}$ for some $a,b,c\in]0,\infty[$, then $P(x,y,z)\iff P(a,b,c)$. This is the symmetry that we are using, since, for any $\{x,y,z\}\subset\mathbb R$, we can re-order the set so that there are real numbers $a\le b\le c$ so that $\{x,y,z\}=\{a,b,c\}$.
On
Here I give a proof by using the standard pqr method.
Proof: Let $p = x+y+z$, $q = xy+yz+zx$ and $r = xyz$.
We will use the following facts (see [1], Facts N12 and N6):
(i) $q^3 + 9r^2 \ge 4pqr$.
(ii) $q^3 \ge 27r^2$.
We need to prove that
$$\frac{pq}{r} \ge 9 \sqrt{\frac{p^2-2q}{q}}$$
or
$$\frac{p^2q^2}{r^2} \ge 81 \frac{p^2-2q}{q}$$
or
$$162qr^2 - (81r^2 - q^3)p^2 \ge 0.$$
There are two possible cases:
1) If $81r^2 - q^3 > 0$: From Fact (i), we have $\frac{q^3+9r^2}{4qr} \ge p$. It suffices to prove that
$$162qr^2 - (81r^2 - q^3)\left(\frac{q^3+9r^2}{4qr}\right)^2 \ge 0$$
or
$$\frac{(q^3 - 9r^2)(q^3 - 27r^2)^2}{16q^2r^2} \ge 0.$$
It is true by using Fact (ii).
2) If $81r^2 - q^3 \le 0$, clearly the inequality is true.
We are done.
Reference:
[1] Zdravko Cvetkovski, "Inequalities Theorems, Techniques and Selected Problems", Ch. 14. https://keoserey.files.wordpress.com/2012/07/zdravko-cvetkovski-inequalities-theorems_-techniques-and-selected-problems.pdf
Remark: In Cvetkovski's book, chapter 14, Page 138, Cvetkovski gave Facts N1 through N13. There is a typo: Fact N10 should be $2q^3 + 9r^2 \ge 7pqr$ (rather than $2p^3 + 9r^2 \ge 7pqr$ which is not true for $a=4, b=3, c=2$).
On
I have a solution using Buffalo way, but it's ugly! I'm sorry about that!
Solution:
Without loss of generality, assume that $x=\min\{x,y,z\}$.
Let $x=a$, $y=a+u$, $z=a+v$ so $a>0$; $u,v \geq 0$
We need to prove: $$(x+y+z)^2 (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2 - 81\frac{(x^2 +y^2 +z^2)}{xy+yz+zx} \geq 0$$
After reduction of many fractions to a common denominator, we need to prove:
$$27a^6(u^2 -uv+v^2)+18a^5 (u+v)^3 +3a^4 (u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 )+2a^3(4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5)+3a^2 uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)+6au^2 v^2(u^3 +4uv(u+v)+v^3)+u^3 v^3 (u+v)^2 \geq 0$$
Because: $u^2 -uv+v^2 \geq 0$; $(u+v)^3 \geq 0$; $u^4 +13u^3 v+78u^2 v^2+13uv^3 +v^4 \geq 0$, $uv(4u^4 +4u^3 v+57u^2 v^2 +4uv^3 +4v^4)$;$(u^3 +4uv(u+v)+v^3)\geq 0$; $u^3 v^3 (u+v)^2 \geq 0$
So it suffices to prove: $4u^5 -7u^4 v+94u^3 v^2 +94u^2 v^3 -7uv^4 +4v^5 \geq 0$
But it's obvious true by AM-GM: $$4u^5+94u^3 v^2 -7u^4 v \geq 2\sqrt{(4u^5).(94u^3 v^2)} - 7u^4 v =(4\sqrt{94}-7)u^4 v >0$$ And $$4v^5 +94u^2 v^3 -7uv^4 \ge 2\sqrt{(4v^5).(94u^2 v^3)} -7uv^4 =(4\sqrt{94}-7)uv^4 >0$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=\frac{uv^2}{w^3}-\sqrt{\frac{3u^2-2v^2}{v^2}}.$$ We see that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ ( https://artofproblemsolving.com/community/c6h278791 )
happens for equality case of two variables.
Since our inequality is homogeneous, we can assume $y=z=1,$ which gives $$(x+2)^2(2x+1)^3\geq81x^2(x^2+2)$$ or $$(x-1)^2(8x^3-21x^2+36x+4)\geq0,$$ which is obvious.