Minimum Value

Expanding, the expression becomes

$16 + 9 + 6*\dfrac{x}{y} + 24*\dfrac{y}{x} = 25 + 6z + \dfrac{24}{z}$

where $z = \dfrac{x}{y}.$ Now since $z \gt 0$ , by the A.M.-G.M. inequality we have that

$\dfrac{6z + \frac{24}{z}}{2} \ge \sqrt{(6z)(\frac{24}{z})} = 12$

$\Longrightarrow 6z + \dfrac{24}{z} \ge 24$ ,

the minimum being achieved when $z = 2$ . Thus the minimum value of the given expression is $25 + 24 = \boxed{49}$ .

By Cauchy Schwarz inequality, this is at least $(\sqrt {2×8}+\sqrt {3×3})^{2}=49$ , and equality can be achieved (e.g. at $x=2, y=1$ ).

Expanding the expression yields :

$25+6*x/y+24*y/x$

Now, let $z :=x/y$ , giving :

$25+6*(z+4/z)$

We need to find the value of $z$ which minimizes $z+4/z$ .

So, let $F(z)=z+4/z$

Looking at an extrema of this function :

$F'(z) = 1-4/z^2 = 0 \iff z^2 = 4$

Now, we require $x,y > 0 \implies z > 0$

So, take $z = 2$

By inspection, one checks that $F(z=2)$ is in fact a minimum and :

$F(z=2) = 4$

Hence, we conclude that that the minimum value of the given expression is :

$25+6*4 = 49$

(Note : one should, to be rigorous, argue that $z = 2$ is indeed attainable by letting $x=2$ and $y=1$ which are acceptable values for $x$ and $y$ )

Using the Python sympy library: - Take partial derivatives of the given expression wrt x and y.

• Solve the resulting pair of equations for x.

• Take note of the fact that only one of the resulting solutions is admissible as a positive number.

• Substitute the admissible expression for x in terms of y back into the original expression.

(I missed seeing what skill one was supposed to apply until just now!)

AM-GM inequality $2x+3y\geq2\sqrt{2x3y}$ and AM-GM inequality $\frac{8}{x}+\frac{3}{y}\geq 2 \sqrt{ \frac{8\cdot 3}{x\cdot y} }$ By multiplying them we obtain that our expression is $\geq 2\cdot 2 \sqrt{2\cdot 8 \cdot 3^2}=48$

(2x+3y)*(8y+3x) 16+9+6x/y+24yx 25+6(x/y+4y/x) jika x/y=p maka persamaaannya(equality) menjadi 25+6(p+4/p) minimum jika p=2 next 25+24=49

Should be 48. Minimum happens when both the factors are minimum which then multiply to 48

Expanding out the expression: $25 + (\frac{6x}{y} + \frac{24y}{x}) = 25 + 6 (\frac{x^2 + 4y^2}{xy})$ Now using the AM-GM inequality: $x^2+4y^2 \geq\ 2\sqrt{x^2. 4y^2} = 4xy$ $\Rightarrow$ $25 + 6 (\frac{x^2 + 4y^2}{xy}) \geq\ 25 + 6(\frac{4xy}{xy}) = 25 + 24 = 49$ $\large Hence \ the \ minimum \ value \ is \ 49$