Find Min of x+y+z for Real x,y,z ≤ 3/2

[anemone]

Find the minimum of $x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ for all positive real $x,\,y$ and $z$ that satisfies the condition $x+y+z\le \dfrac{3}{2}$.

 

[Greg]

Find the minimum of $x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ for all positive real $x,\,y$ and $z$ that satisfies the condition $x+y+z\le \dfrac{3}{2}$.

Spoiler

$$x+y+z+\dfrac1x+\dfrac1y+\dfrac1z\ge\dfrac32+a$$ where $a$ is some non-negative constant.

$$x+y+z-\dfrac32+\dfrac1x+\dfrac1y+\dfrac1z\ge a$$

$$b=x+y+z-\dfrac32\implies-\dfrac32\lt b\le0$$

$$\dfrac1x+\dfrac1y+\dfrac1z\ge a+|b|$$

Applying the AM-GM inequality, we have $$\dfrac{\dfrac1x+\dfrac1y+\dfrac1z}{3}\ge\left(\dfrac{1}{xyz}\right)^{1/3}$$
hence $$\dfrac1x+\dfrac1y+\dfrac1z$$ is at a minimum when $x=y=z=\dfrac12$.

From this, $b=0$ and $a=6$, so the desired minimum is $6+\dfrac32=\dfrac{15}{2}$.

 

[anemone]

Great job, greg1313!:)

My solution:

Spoiler

Note that from AM-GM inequality we get $3\sqrt[3]{xyz}\le x+y+z\le \dfrac{3}{2}$, which translates to $\sqrt[3]{xyz}\le \dfrac{1}{2}$, we can conclude $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \dfrac{3}{\sqrt[3]{xyz}}=\dfrac{3}{\dfrac{1}{2}}=6$.

By AM-HM we have $x+y+z\ge \dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}$ so the intended expression becomes

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

Since $x,\,y$ and $z$ are positive real numbers and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\ge 2\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\right)}=6$, we can conclude by now that

$\begin{align*}x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}&\ge \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\\&\ge 6+\dfrac{9}{6}\\&\ge \dfrac{15}{2}\end{align*}$

Equality occurs when $\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=\dfrac{1}{2}$.

 

[Albert1]

My solution:

Spoiler

Note that from AM-GM inequality we get $3\sqrt[3]{xyz}\le x+y+z\le \dfrac{3}{2}$, which translates to $\sqrt[3]{xyz}\le \dfrac{1}{2}$, we can conclude $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \dfrac{3}{\sqrt[3]{xyz}}=\dfrac{3}{\dfrac{1}{2}}=6$.

By AM-HM we have $x+y+z\ge \dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}$ so the intended expression becomes

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

Since $x,\,y$ and $z$ are positive real numbers and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\ge 2\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\right)}=6---(A)$, we can conclude by now that

$\begin{align*}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{9}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\\\ge 6+\dfrac{9}{6}\ge \dfrac{15}{2}---(B)\end{align*}$

Equality occurs when $\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=\dfrac{1}{2}$.

I know the answer is correct but---
step from (A)to (B) can you explain ?

 

[anemone]

My solution:
I know the answer is correct but---
step from (A)to (B) can you explain ?

Sorry Albert that I wasn't clear...

Spoiler

I will explain using the example function of $f(a)=a+\dfrac{9}{a}$, note that we can apply the AM-GM inequality to the function of $f$ when $a$ is positive to get $f(a)=a+\dfrac{9}{a}\ge 2(3)=6$. Equality happens when $a=3$.

And we know the function of $f$ is an increasing function beyond $a=3$. Since $a\ge 6$, the minimum of $f$ would then occur at $a=6$, so we get $f(a)_{min}=6+\dfrac{9}{6}=\dfrac{15}{2}$.

 

[Albert1]

Sorry Albert that I wasn't clear...

Spoiler

I will explain using the example function of $f(a)=a+\dfrac{9}{a}$, note that we can apply the AM-GM inequality to the function of $f$ when $a$ is positive to get $f(a)=a+\dfrac{9}{a}\ge 2(3)=6$. Equality happens when $a=3$.

And we know the function of $f$ is an increasing function beyond $a=3$. Since $a\ge 6$, the minimum of $f$ would then occur at $a=6$, so we get $f(a)_{min}=6+\dfrac{9}{6}=\dfrac{15}{2}$.

many thanks! now I got it