Optimizing the "Cool Function"

First we need to show that $r+\frac{1}{r}\ge2$ for any positive $r$ . To show this, we first take a case of the trivial inequality; that is, $(r-1)^2 \ge 0$ . Expanding this yields $r^2-2r+1\ge0$ . Adding $2r$ to both sides gives us $r^2+1 \ge 2r$ . Since $r$ was assumed to be positive, we divide through by $r$ to get $r+\frac{1}{r} \ge 2$ .

Now consider the Cool Function $f$ . Note that $f$ can be simplified into

$f(x,y,z,t)=\frac{x^2+2x+1}{x}\cdot\frac{y^2+2x+1}{y}\cdot\frac{z^2+z+1}{z}\cdot\frac{t^2+t+1}{t}$

Consider the fraction $\frac{x^2+2x+1}{x}=x+2+\frac{1}{x}$ . From the inequality we proved earlier, we have that $x+\frac{1}{x} \ge 2$ , so that $x+2+\frac{1}{x} \ge 4$ . By the same logic, we find $\frac{y^2+2x+1}{y} \ge 4$ and $\frac{z^2+z+1}{z},\frac{t^2+t+1}{t} \ge 3$ . Thus we have that $f(x,y,z,t)\ge4\cdot4\cdot3\cdot3=\boxed{144}$

We rewrite $f(x, y, z, t)$ as $\left(x + 2 + \dfrac1x\right)\left(y + 2 + \dfrac1y\right)\left(z + 1 + \dfrac1z\right)\left(t + 1 + \dfrac1t\right)$ Notice $\dfrac{x + \dfrac{1}{x}}{2} \ge \sqrt{x \cdot \dfrac{1}{x}} \ge 1$ , so $x + \dfrac{1}{x} \ge 2$ . Similarly, $y + \dfrac{1}{y} \ge 2$ , $z + \dfrac{1}{z} \ge 2$ , and $t + \dfrac{1}{t} \ge 2$ . Therefore, we have $\begin{aligned}f(x,y,z,t) &=& \left(x + 2 + \dfrac1x\right)\left(y + 2 + \dfrac1y\right)\left(z + 1 + \dfrac1z\right)\left(t + 1 + \dfrac1t\right)\\ &\ge& (2 + 2)(2 + 2)(1 + 2)(1 + 2) \\&\ge& 144\end{aligned}$ Equality holds when $x=y=z=t=1$ . $\blacksquare$

By inspecting the Cool Function $f$ , we note that it can be expressed as a product of four other single-variable functions (of $x,y, z$ and $t$ .) That is, $f(x,y,z,t)=A(x)\cdot B(y)\cdot C(z) \cdot D(t)$ . Hence, $f$ is minimum whenever $A, B,C,$ and $D$ are all minimum, i.e, we optimize each function independently.

Since functions $A$ and $B$ are of the same form, as well as the pair of functions $C$ and $D$ , we optimize just one of them (the minimum of the other is equivalent to that of its pair.)

Using the concept of derivatives , we optimize functions $A$ and $C$ . We also note that all the functions are only defined from the positive reals.

$\frac{dA}{dx}=\frac{d}{dx}(\frac{x^2+2x+1}{x})=\frac{x^2-1}{x^2}=0 \\ \frac{dC}{dz}=\frac{d}{dz}(\frac{z^2+z+1}{z})=1-\frac{1}{z^2}=0$

Therefore, the minimum value of $f(x,y,z,t)$ occurs at $x=y=z=t=1$ . Evaluating $f(1,1,1,1)$ gives $\boxed{144}$ .

The function f(x,y,z,t) is actually multiplying four different functions together:

$g(x) = \frac {x^2 + 2x + 1}{x}$

$g(y) = \frac {y^2 + 2y + 1}{y}$

$h(z) = \frac {z^2 + z + 1}{z}$

$h(t) = \frac {t^2 + t + 1}{t}$

Each of those functions are from the positive reals, and it's easy to see that the functions themselves are always equal to positive reals. (because the numerator is always positive, and so is the denominator, since x, y, z, and t are all positive)

Therefore, if I want to find the minimum value for $f(x)$ , I need to find the minimum value for each of the other four functions, and then multiply them together! All that is needed for this is a bit of calculus. The first two functions are actually the same - they only have different variables: x and y. Let's start with that.

First, I'd like to simplify the fraction a bit:

$g(x) = \frac {x^2 + 2x + 1}{x} = x + 2 + \frac {1}{x}$

To find the minimum value we have to differentiate the function, and solve for when the slope is zero:

$g'(x) = 1 - \frac {1}{x^2} = 0$

If we solve for x:

$\frac {1}{x^2} = 1$

$1 = x^2$ (here I multiplied by $x^2$ )

$x = \pm 1$ ---Note that since $x$ is positive (this is a given), the only value it can be is:

$x = 1$ ---Just to make sure, let's substitute this in $g''(x)$ so that we know it's a minimum!

$g''(x) = \frac {2}{x^3}$

$g''(1)= \frac {2}{1^3} = \frac {2}{1} = 2 > 0$ --- The value of $g''(x)$ is positive, so it is a minimum.

Now we can substitute this in $g(x)$ to find the minimum value. Since we also have the same function in terms of y: $g(y)$ , then we can substitute the same value in that function too:

$g(x)_{min} = g(y)_{min} = g(1) = \frac {1^2 + 2 \cdot 1 + 1}{1} = 4$

Let's do the same for $h(z)$ and $h(t)$ : Differentiate and find the minimum value. The process is the same:

$h(z) = \frac {z^2 + z + 1}{z} = z + 1 + \frac {1}{z}$

Now we differentiate:

$h'(z) = 1 - \frac {1}{z^2} = 0$

If we solve for z:

$\frac {1}{z^2} = 1$

$1 = z^2$ (here I multiplied by $z^2$ )

$z = \pm 1$ ---Note that since $z$ is positive, the only value it can be is:

$z = 1$ ---Just to make sure, let's substitute this in $h''(z)$ so that we know it's a minimum!

$h''(z) = \frac {2}{z^3}$

$h''(1)= \frac {2}{1^3} = \frac {2}{1} = 2 > 0$ --- The value of $h''(z)$ is positive, so it is a minimum.

Now we can substitute this in $h(z)$ to find the minimum value. Since we also have the same function in terms of t: $h(t)$ , then we can substitute the same value in that function too:

$h(z)_{min} = h(t)_{min} = h(1) = \frac {1^2 + 1 + 1}{1} = 3$

Everything left is to multiply the minimum values of all of those functions to get the minimum value for $f(x,y,z,t)$ .

Ready? (drumroll please....) Here goes....

$f(x,y,z,t)_{min} = g(x)_{min} \cdot g(y)_{min} \cdot h(z)_{min} \cdot h(t)_{min} = 4 \cdot 4 \cdot 3 \cdot 3 = 144$ .

And so, the minimum value of the function is:

$\boxed{f(x,y,z,t)_{min} = 144}$

f(x,y,z,t) = [g(x)+1]·[g(y)+1]·g(z)·g(t) . . . where g(x) = x + 1/x + 1

But for x>0, Min(x + 1/x) = 2 (easily verified by calculus)

∴ Min[g(z)] = 2+1 = 3, and

Min(f(x,y,z,t)) = Min[g(x)+1]·Min[g(y)+1]·Min[g(z)]·Min[g(t)] = 4·4·3·3 = 12² = 144

We have

(x^2 + 2x + 1)/x = x + 2 + 1/x >= 4 ......(i)

Reason: Using AM-GM inequality:

x + 1/x >= 2

Similarly

(y^2 + 2y + 1)/y = y + 2 + 1/y >= 4 ........(ii)

Also,

(z^2 + z + 1)/z = z + 1 + 1/z >= 3 ...........(iii)

because of the same reason.

Similarly

(t^2 + t + 1)/t = t + 1 + 1/t >= 3 ...........(iv)

Now, multiplying the inequalities in (i), (ii), (iii) and (iv) we get that

f(x,y,z,t) >= 144

Hence, the minimum value of f(x,y,z,t) is 144.

$f(x,y,z)=(x+1/x+2)(y+1/y+2)+(z+1/z+1)(t+1/t+1)$ . Now, since $x,y,z,t\in {R}$ for each of them the inequality $a+1/a\ge 2$ holds true. Hence $f(x,y,z)\ge 4\cdot 4\cdot 3\cdot 3=144$ . Also, the function attains this value when $x=y=z=t=1$ .

Simplifying x^2+2x+1 terms to x+1/x +2 and the other terms to x^3-1/x-1 and using maxima minima gives a direct answer