What is the minimum value of the expression $$xy+yz+zx+\frac 1x+\frac 2y+\frac 5z$$where $x, y, z$ are positive real numbers?
Problem
Source: Turkey TST 2022 P7 Day 3
Tags: inequalities, algebra
BarisKoyuncu
13.03.2022 14:49
$3\sqrt[3]{36}$
$(x,y,z)=\left(\frac{\sqrt[3]{6}}3, \frac{\sqrt[3]{6}}2, \sqrt[3]{6}\right)$
By AM-GM inequality,
$$xy+\frac{1}{3x}+\frac{1}{2y}\ge \frac{\sqrt[3]{36}}2$$$$yz+\frac{3}{2y}+\frac{3}{z}\ge \frac{3\sqrt[3]{36}}2$$$$zx+\frac{2}{z}+\frac{2}{3x}\ge \sqrt[3]{36}$$Summing up all these inequalities gives us the desired result.
We are trying to find real numbers $0<a<1, 0<b<2, 0<c<5$ such that there exist real numbers $x,y,z$ satisfying
$$xy=\frac{a}{x}=\frac{b}{y}$$$$yz=\frac{2-b}{y}=\frac{c}{z}$$$$zx=\frac{5-c}{z}=\frac{1-a}{x}$$Then we can apply AM-GM, and finish the proof
(Notice that this gives us both a lower bound and an equality case)
Sum up the first and the last equations. We need to have $xy+zx=\frac{a}{x}+\frac{1-a}{x}=\frac 1x$. Hence, $x^2(y+z)=1$. Similarly, one can find that $y^2(x+z)=2$ and $z^2(x+y)=5$. Summing up these $3$ equations gives us $\sum_{sym} x^2y=8$ and then multiplying them gives $(xyz)^2(8+2xyz)=10\Rightarrow xyz=1$.
Then, $y=x^2(y^2+yz)=x^2y^2+x$ and $2x=y^2(x^2+xz)=x^2y^2+y$. Summing up these equalities gives $x=2x^2y^2\Rightarrow xy^2=\frac 12$. Therefore, $b=xy^2=\frac 12$. Similarly, one can find that $a=\frac 13$ and $c=3$. The rest is shown above.
sqing
13.03.2022 16:22
BarisKoyuncu wrote:
Answer$3\sqrt[3]{36}$
Equality case$(x,y,z)=\left(\frac{\sqrt[3]{6}}3, \frac{\sqrt[3]{6}}2, \sqrt[3]{6}\right)$
ProofBy AM-GM inequality,
$$xy+\frac{1}{3x}+\frac{1}{2y}\ge \frac{\sqrt[3]{36}}2$$$$yz+\frac{3}{2y}+\frac{3}{z}\ge \frac{3\sqrt[3]{36}}2$$$$zx+\frac{2}{z}+\frac{2}{3x}\ge \sqrt[3]{36}$$Summing up all these inequalities gives us the desired result.
MotivationWe are trying to find real numbers $0<a<1, 0<b<2, 0<c<5$ such that there exist real numbers $x,y,z$ satisfying
$$xy=\frac{a}{x}=\frac{b}{y}$$$$yz=\frac{2-b}{y}=\frac{c}{z}$$$$zx=\frac{5-c}{z}=\frac{1-a}{x}$$Then we can apply AM-GM, and finish the proof
(Notice that this gives us both a lower bound and an equality case)
Sum up the first and the last equations. We need to have $xy+zx=\frac{a}{x}+\frac{1-a}{x}=\frac 1x$. Hence, $x^2(y+z)=1$. Similarly, one can find that $y^2(x+z)=2$ and $z^2(x+y)=5$. Summing up these $3$ equations gives us $\sum_{sym} x^2y=8$ and then multiplying them gives $(xyz)^2(8+2xyz)=10\Rightarrow xyz=1$.
Then, $y=x^2(y^2+yz)=x^2y^2+x$ and $2x=y^2(x^2+xz)=x^2y^2+y$. Summing up these equalities gives $x=2x^2y^2\Rightarrow xy^2=\frac 12$. Therefore, $b=xy^2=\frac 12$. Similarly, one can find that $a=\frac 13$ and $c=3$. The rest is shown above.
CahitArf
20.03.2022 23:08
I have a bit longer but different proof.
Attachments:
Turkey TST P7.docx (18kb)
sqing
21.03.2022 03:30
Let $x, y, z$ are positive real numbers. Prove that$$ xy+yz+ zx+\frac{1}{x}+\frac{3}{y}+\frac{3}{z} \geq 6\sqrt[3]{4}$$$$ xy+2yz+ zx+\frac{1}{x}+\frac{3}{y}+\frac{3}{z} \geq \frac{15\sqrt[3]{3}}{2}$$$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+ab+bc+ca \geq \frac{9}{\sqrt[3]{4}}$$h h CahitArf wrote: I have a bit longer but different proof.
Attachments:
sqing
30.05.2022 09:22
Let $x, y, z$ are positive real numbers. Prove that $$ xy+2yz+2zx+\frac{2}{x}+\frac{2}{y}+\frac{1}{z} \geq 9$$$$ xy+zx+\frac{5}{x}+\frac{2}{y}+\frac{2}{z} \geq 6\sqrt[3]{5}$$$$ x+3y+3z+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \geq 6\sqrt[3]{4}$$ BarisKoyuncu wrote: What is the minimum value of the expression $$xy+yz+zx+\frac 1x+\frac 2y+\frac 5z$$where $x, y, z$ are positive real numbers? $$xy+yz+zx+\frac 1x+\frac 2y+\frac 5z\geq 3\sqrt[3]{36}$$$$ xy+yz+ zx+\frac{1}{x}+\frac{3}{y}+\frac{3}{z} \geq 6\sqrt[3]{4}$$
Attachments:
ehuseyinyigit
31.10.2023 22:54
This problem has a good general one however I will give the minimum.
ehuseyinyigit
31.10.2023 22:55
Generalization 1 Let $x,y,z,\lambda,a_{1},b_{1}$ be positive real numbers. Then prove that $$\lambda \left(xy+yz+xz\right) + \dfrac{a_{1}\left(b_{1}+1\right)}{b_{1}x}+\dfrac{b_{1}\left(a_{1}+1\right)}{a_{1}y}+\dfrac{a_{1}+b_{1}}{a_{1}b_{1}z} \geq 3\sqrt[3]{\lambda a_{1}b_{1}}\left(\dfrac{a_{1}+b_{1}+a_{1}b_{1}}{a_{1}b_{1}}\right)$$
ehuseyinyigit
05.11.2023 21:22
The original problem and Generalization 1 are special cases of the following generalization where $$n=3; \lambda =1, a_{1}=\dfrac{1}{3}, a_{2}=b_{1}=\dfrac{1}{2}$$
ehuseyinyigit
05.11.2023 21:23
Generalization 2 Let $x_{1},x_{2},\cdots,x_{n},\lambda ,a_{1},a_{2},\cdots,a_{n-1}$ be positive reals ($n\geq 3$). Then prove that $$\sum_{j=1}^{n-1}{\left(\dfrac{a_{j}\left(1+\sum_{p=1}^{n-1}{\left(\dfrac{1}{a_{p}}\right)-\dfrac{1}{a_{j}}}\right)}{x_{j}}\right)}+\dfrac{\sum_{k=1}^{n-1}{\left(\dfrac{1}{a_{k}}\right)}}{x_{n}}\geq n\sqrt[n]{\lambda \prod{a_{1}}}\left(1+\sum_{p=1}^{n-1}{\dfrac{1}{a_{p}}}\right)$$