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

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Minimum
AI Thread Summary
The discussion revolves around finding the minimum value of the expression x+y+z+1/x+1/y+1/z under the constraint that x+y+z ≤ 3/2 for positive real numbers x, y, and z. Participants share their solutions and seek clarification on specific steps in the problem-solving process. One user expresses confidence in their answer but requests an explanation for a transition in their solution. The conversation highlights the collaborative nature of mathematical problem-solving, with users helping each other understand complex steps. Ultimately, the discussion emphasizes the importance of clear communication in resolving mathematical queries.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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 $x+y+z\le \dfrac{3}{2}$.
 
Mathematics news on Phys.org
anemone said:
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 $x+y+z\le \dfrac{3}{2}$.

$$x+y+z+\dfrac1x+\dfrac1y+\dfrac1z\ge\dfrac32+a$$ 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 $6+\dfrac32=\dfrac{15}{2}$.
 
Great job, greg1313!:)

My solution:

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 $\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=\dfrac{1}{2}$.
 
My solution:
anemone said:
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 ?
 
Last edited:
Albert said:
My solution:
I know the answer is correct but---
step from (A)to (B) can you explain ?

Sorry Albert that I wasn't clear...

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 $f(a)_{min}=6+\dfrac{9}{6}=\dfrac{15}{2}$.
 
anemone said:
Sorry Albert that I wasn't clear...

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
 
Last edited:
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top