Solve Let \(a\), \(b\), and \(c\) be positive real numbers such that \(a + b + c = 3\).Prove the following inequality:\(a3b+c+b3c+a+c3a+b (3)/(2)\)

Given $a+b+c=3$, we can express the denominators as:
$b+c = 3-a$
$c+a = 3-b$
$a+b = 3-c$
Substituting these into the inequality, we get:

$$\frac{a^{3}}{3-a}+\frac{b^{3}}{3-b}+\frac{c^{3}}{3-c}\ge \frac{3}{2}$$

Let $f(x) = \frac{x^3}{3-x}$ for $x \in (0, 3)$. Since $a, b, c$ are positive and sum to 3, each must be less than 3.
We need to find the second derivative of $f(x)$ to check for convexity.
First, find the first derivative $f'(x)$:

$$f'(x) = \frac{d}{dx}\left(\frac{x^3}{3-x}\right) = \frac{3x^2(3-x) - x^3(-1)}{(3-x)^2} = \frac{9x^2 - 3x^3 + x^3}{(3-x)^2} = \frac{9x^2 - 2x^3}{(3-x)^2}$$

Next, find the second derivative $f''(x)$:

$$f''(x) = \frac{d}{dx}\left(\frac{9x^2 - 2x^3}{(3-x)^2}\right)$$

Using the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$:
Let $u = 9x^2 - 2x^3 \implies u' = 18x - 6x^2$
Let $v = (3-x)^2 \implies v' = 2(3-x)(-1) = -2(3-x)$

$$f''(x) = \frac{(18x - 6x^2)(3-x)^2 - (9x^2 - 2x^3)(-2(3-x))}{(3-x)^4}$$

Factor out $(3-x)$ from the numerator:

$$f''(x) = \frac{(3-x)\left[(18x - 6x^2)(3-x) + 2(9x^2 - 2x^3)\right]}{(3-x)^4}$$ $$f''(x) = \frac{(18x - 6x^2)(3-x) + 2(9x^2 - 2x^3)}{(3-x)^3}$$

Expand the numerator:

$$f''(x) = \frac{(54x - 18x^2 - 18x^2 + 6x^3) + (18x^2 - 4x^3)}{(3-x)^3}$$ $$f''(x) = \frac{2x^3 - 18x^2 + 54x}{(3-x)^3}$$

Factor out $2x$ from the numerator:

$$f''(x) = \frac{2x(x^2 - 9x + 27)}{(3-x)^3}$$

For $x \in (0, 3)$:

• $2x > 0$
• $(3-x)^3 > 0$
• For the quadratic $x^2 - 9x + 27$, its discriminant is $\Delta = (-9)^2 - 4(1)(27) = 81 - 108 = -27$. Since $\Delta < 0$ and the leading coefficient (1) is positive, $x^2 - 9x + 27 > 0$ for all real $x$. Therefore, $f''(x) > 0$ for all $x \in (0, 3)$. This means $f(x)$ is a convex function on the interval $(0, 3)$.

Since $f(x)$ is convex, by Jensen's inequality:

$$\frac{f(a)+f(b)+f(c)}{3} \ge f\left(\frac{a+b+c}{3}\right)$$

Substitute $f(x) = \frac{x^3}{3-x}$ and the given condition $a+b+c=3$:

$$\frac{1}{3}\left(\frac{a^{3}}{3-a}+\frac{b^{3}}{3-b}+\frac{c^{3}}{3-c}\right) \ge \frac{\left(\frac{a+b+c}{3}\right)^3}{3-\left(\frac{a+b+c}{3}\right)}$$ $$\frac{1}{3}\left(\frac{a^{3}}{3-a}+\frac{b^{3}}{3-b}+\frac{c^{3}}{3-c}\right) \ge \frac{\left(\frac{3}{3}\right)^3}{3-\left(\frac{3}{3}\right)}$$ $$\frac{1}{3}\left(\frac{a^{3}}{3-a}+\frac{b^{3}}{3-b}+\frac{c^{3}}{3-c}\right) \ge \frac{1^3}{3-1}$$ $$\frac{1}{3}\left(\frac{a^{3}}{3-a}+\frac{b^{3}}{3-b}+\frac{c^{3}}{3-c}\right) \ge \frac{1}{2}$$

$$\frac{a^{3}}{b+c}+\frac{b^{3}}{c+a}+\frac{c^{3}}{a+b} \ge \frac{3}{2}$$

The inequality is proven.