- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 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 $x+y+z\le \dfrac{3}{2}$.
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 $x+y+z\le \dfrac{3}{2}$.
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 $\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}=\dfrac{1}{2}$.
Albert said:My solution:
I know the answer is correct but---
step from (A)to (B) can you explain ?
many thanks! now I got itanemone 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 $f(a)_{min}=6+\dfrac{9}{6}=\dfrac{15}{2}$.
This phrase is asking for the smallest possible value of the sum of three real numbers (x, y, and z) that are all less than or equal to 3/2. In other words, you are looking for the minimum possible value of x+y+z when x, y, and z are restricted to a maximum value of 3/2.
The most efficient way to solve this problem is by using the method of Lagrange multipliers. This involves finding the critical points of the function f(x,y,z) = x+y+z subject to the constraint g(x,y,z) = x+y+z-3/2 = 0. The minimum value will be at one of these critical points.
Yes, this problem can be solved using calculus. As mentioned before, the method of Lagrange multipliers involves using partial derivatives and setting them equal to each other to find the critical points. From there, we can use the second derivative test to determine if the critical point is a minimum or maximum.
Yes, there are alternative ways to solve this problem without using calculus. One approach is by using the concept of inequalities. We can set up the problem as a system of linear inequalities and use techniques such as graphing or substitution to find the minimum value.
Yes, the solution to this problem will be unique. The method of Lagrange multipliers guarantees that the critical point we find will be a global minimum, meaning it will be the smallest possible value for the sum of x, y, and z. However, there may be multiple combinations of x, y, and z that result in this minimum value.