Least integral value!

By A.M. - G.M. :

$x + \frac{25}{x} \geq 2 \cdot \sqrt{25} = 10$

Consider $(x - 5)^{2} \ge 0$

$x^{2} - 10x + 25 \ge 0$

$x^2 + 25 \ge 10x$

As x is non-zero, $x + \frac{25}{x} \ge 10$

Note:

$x+\frac{25}{x}=(\frac{x^{2} +25}{x})$ $=(\frac{x^{2} -10x +25}{x}) +\frac{10x}{x}$ Completing the square $x+\frac{25}{x}=\frac{1}{x} (x-5)^{2} +10$ Since x is positive the least value of $\frac{1}{x} (x-5)^{2}$ is 0.

Therefore, the least integral value of $\frac{1}{x} (x-5)^{2} +10$ is 10.

You can also use the method of differentiation but in this kind of problem, much better to know the basic of algebra...

Differentiating the term we get $1x^{ 0 }-1(\frac { 25 }{ { x }^{ 2 } } )$ .
Since the slope at the minimum value is zero
$\Rightarrow 1-\frac { 25 }{ { x }^{ 2 } } =0\\ \Rightarrow 1\quad =\quad \frac { 25 }{ { x }^{ 2 } } \\ \Rightarrow x\quad =\quad \pm 5$

Hence the positive value is $\boxed {5}$

I considered the equation as f ( x ) and calculated d/d x f ( x ) equating it to zero.

It gave me, ( x )^2 = 25

Thus, at x = 5 , we can get minimum value of function f ( x ) , which is nothing but 10.

y=(x^2+25)/x y' = x^2 - 25 = 0 Since X is positive then it must be +5 So X + 25/X = 10

$x+\frac { 25 }{ x } =y,\quad y\in \quad R.\quad It\quad is\quad obvious\quad that\quad x\neq 0,\quad thus\quad { x }^{ 2 }-xy+25=0.\\ \Delta ={ y }^{ 2 }-4.25.\quad For\quad it\quad to\quad have\quad real\quad solutions,\quad we\quad set\quad \Delta \ge 0\\ So\quad { y }^{ 2 }\ge 100\Longleftrightarrow \left| y \right| \ge 10.\quad Since\quad x\quad is\quad positive,\quad y\quad is\quad also\quad positive.\\ The\quad least\quad value\quad y\quad can\quad get\quad is\quad 10,\quad so\quad this\quad is\quad the\quad answer,\quad for\quad x=5.$

Begin by writing:

$x + \frac{25}{x} = \frac{x^2 + 25}{x} = \frac{x^2 + 5^2}{x}$

Then we define a variable Q to be the lowest value of the fraction

$Q = \frac{x^2 + 5^2}{x} \Rightarrow Qx = x^2 + 5^2 \Rightarrow x^2 - Qx + 5^2 = 0$

Then we use the quadratic formula to get

$x = \frac{-(-Q) \pm \sqrt{Q - 4(5^2)(1)}}{2 * 1} \Rightarrow Q^2 - 100 \geq 0$

We also get that $\frac{-(-Q) \pm \sqrt{Q - 4(5^2)(1)}}{2 * 1} \geq 0$

If we continue with our argument that $Q^2 - 100 \geq 0 \Rightarrow$ $Q^2 \geq 100 \Rightarrow$ $Q \geq \sqrt{100} = 10$

Here we see that the lowerst value of Q is 10, but we need to see that the other constraint is true aswell $\frac{-(-10) \pm \sqrt{10^2 - 4(5^2)(1)}}{2 * 1} =$ $\frac{10 \pm \sqrt{100 - 100}}{2} =$ $\frac{10}{2} =$ $5 \geq 0$

Therefore the lowerst value for the expression $x + \frac{25}{x}$ will be when x = 10.

Get the first derivative and equate to $0$ , $x$ must be $5$ .

5 + 25/5 = 5 + 5 = 10