Animated Solution for Mathematics - Differentiation: If $a{x}^2+\frac{b}{x}\ge c$ for all positive $x$ where $a>0$ and $b>0$ show that $27a{b}^2\ge 4c^3$.

Imagine you are standing on a landscape defined by the function $f(x)=a{x}^2+\frac{b}{x}$. As $x$ approaches zero, the $\frac{b}{x}$ term dominates, shooting the function up toward infinity.

As $x$ grows very large, the $a{x}^2$ term takes over, again driving the function toward infinity. Somewhere in between, there is a valley—a global minimum.

The problem asks us to prove that $27a{b}^2\ge 4c^3$ given that $f(x)\ge c$ for all $x>0$. This is not just an algebraic exercise; it is a fundamental truth about balancing growth and decay.

Finding the Valley

To find the lowest point of this landscape, we turn to the power of calculus. We define our function as $f(x)=a{x}^2+\frac{b}{x}$.

To find the critical point where the slope is zero, we calculate the first derivative:

Setting $f^{\prime }(x)=0$ gives us the equation $2ax=\frac{b}{x^2}$. Multiplying both sides by $x^2$ yields $2a{x}^3=b$, which simplifies to $x^3=\frac{b}{2a}$.

Thus, our critical point is $x_{min}={\left(\frac{b}{2a}\right)}^{1/3}$. This is the exact point where the function reaches its absolute minimum.

The Minimum Value

Now, we must find the value of the function at this critical point. We substitute $x_{min}$ back into the original function:

Simplifying the first term, we get $a\cdot \frac{b^{2/3}}{(2a{)}^{2/3}}=\frac{a^{1/3}{b}^{2/3}}{2^{2/3}}$. The second term simplifies to $b\cdot {\left(\frac{2a}{b}\right)}^{1/3}=b^{2/3}\cdot 2^{1/3}\cdot a^{1/3}$.

Combining these, we factor out $a^{1/3}{b}^{2/3}$ to get:

Finding a common denominator, we get:

The Final Proof

We know that for the inequality $f(x)\ge c$ to hold for all $x$, the minimum value must be at least $c$. So, $\frac{3a^{1/3}{b}^{2/3}}{2^{2/3}}\ge c$.

To clear the fractional powers, we cube both sides:

This simplifies to $27\cdot \frac{a{b}^2}{4}\ge c^3$. Multiplying by 4, we arrive at the beautiful result: $27a{b}^2\ge 4c^3$.

AM-GM

While calculus is robust, the AM-GM inequality offers a shortcut. By splitting $\frac{b}{x}$ into $\frac{b}{2x}+\frac{b}{2x}$, we can apply the inequality to the three terms $a{x}^2$, $\frac{b}{2x}$, and $\frac{b}{2x}$.

The product of these terms is:

The AM-GM inequality states that the arithmetic mean is greater than or equal to the geometric mean, leading us directly to $f(x)\ge 3{\left(\frac{a{b}^2}{4}\right)}^{1/3}$. This is the same minimum value we found with calculus. Both paths lead to the same truth.