Minimization

Let $f(x) = \dfrac{x^n}{3-x}$ . Then the expression is equivalent to $\dfrac{a^n}{3-a} + \dfrac{b^n}{3-b} + \dfrac{c^n}{3-c} = f(a) + f(b) + f(c)$ .

Since $a,b,c$ must be positive numbers that sum to $3$ , we must have $0 < a,b,c < 3$ .

I am going to use Jensen's Inequality to find the minimum. However, in order to use it, I will first prove that $f$ is concave upward for $0 < x < 3$ :

$f'(x) = \dfrac{(3-x)nx^{n-1} - (-1)x^n}{(3-x)^2} = \dfrac{nx^{n-1}}{3-x} + \dfrac{x^n}{(3-x)^2}$ $f''(x) = n\left(\dfrac{(n-1)x^{n-2}}{3-x} + \dfrac{x^{n-1}}{(3-x)^2}\right) + \dfrac{(3-x)^2 nx^{n-1} - (-2)(3-x)x^n}{(3-x)^4}$ $f''(x) = \dfrac{n(n-1)x^{n-2}}{3-x} + \dfrac{2nx^{n-1}}{(3-x)^2} + \dfrac{2x^n}{(3-x)^3}$

Clearly, if $0 < x < 3$ and $n \ge 1$ , then the first term in that expression is nonnegative, and the other two are positive. Hence $f''(x) > 0$ , so $f(x)$ is concave upward for $0 < x < 3$ .

Now Jensen's Inequality tells us:

$\dfrac{f(a) + f(b) + f(c)}{3} \ge f\left(\dfrac{a+b+c}{3}\right)$ $f(a) + f(b) + f(c) \ge 3f(1) = 3\left(\dfrac{1^n}{3-1}\right) = \dfrac{3}{2}$

If $a = b = c = 1$ , we actually get this answer. Therefore the minimum value is $\boxed{\dfrac{3}{2}}$ .

Applying $\displaystyle AM-GM$ inequality:

$\displaystyle \frac{a^n}{b+c} + \frac{b^n}{a+c} + \frac{c^n}{a+b} \geq 3 \times \sqrt[3]{\frac{a^n \times b^n \times c^n}{(b+c)(a+c)(a+b)}}$

Therefore the minimum value of the given expression(C) can be thus written as:

$\displaystyle \Rightarrow C = 3 \times \sqrt[3]{\frac{(abc)^{n}}{(b+c)(a+c)(a+b)}}$

Condition for equality is:

$\displaystyle \frac{a^n}{b+c} = \frac{b^n}{a+c} = \frac{c^n}{a+b}$

There is a short method and another long one to arrive at the conclusion that this condition finally boils down to $\displaystyle a=b=c$ .

Short method: From the symmetry in the formula of condition for equality, we conclude that the condition can be simply stated as $\displaystyle a=b=c$ .

Longer method:

Consider: $\displaystyle \frac{a^n}{b^n} = \frac{b+c}{a+c}$

$\displaystyle \Big(\frac{a}{b} \Big)^n = \frac{(3-a) }{(3-b)}$

$\displaystyle \Big(\frac{a}{b}\Big)^n \times (3-b) = (3-a)$

Dividing on both sides by $\displaystyle b$ and substituting $\displaystyle t=\frac{a}{b}$ ,

$\displaystyle t^n \times (\frac{3}{b} - 1) = (\frac{3}{b} - t)$

The only real root of this equation is $\displaystyle t=1$ , and the proof for which by induction is:

$\displaystyle \text{For } n=1 :$

$\displaystyle \Rightarrow t \times (\frac{3}{b} - 1) = (\frac{3}{b} - t)$

$\displaystyle \Rightarrow (t-1) \times (\frac{3}{b}) = 0$

$\displaystyle \Rightarrow \boxed{t=1}$

Now we assume that $\displaystyle t=1$ is the only real root for $\displaystyle n = k$ , then

$\displaystyle \Rightarrow t^k \times (\frac{3}{b} - 1) = (\frac{3}{b} - t) \ldots \text{equation}\{1\}$

$\displaystyle \text{For } n = (k + 1) :$

$\displaystyle \Rightarrow t^{k + 1} \times (\frac{3}{b} - 1) = (\frac{3}{b} - t)$

$\displaystyle \Rightarrow t \times (t^k \times (\frac{3}{b} - 1) ) =(\frac{3}{b} - t) \ldots \text{equation}\{2\}$

From equation{1} and {2} we get :

$\displaystyle \Rightarrow t \times (\frac{3}{b} - t) = (\frac{3}{b} - t)$

$\displaystyle \Rightarrow t=1 \text{ or } t=\frac{3}{b}$

But $\displaystyle t \neq \frac{3}{b}$ as that would imply $\displaystyle a=3$ and $\displaystyle b=c=0$ , but $a, b, c$ are positive real numbers and cannot be equal to zero.

Thus, $\displaystyle t=1$ is the only real root and so $\displaystyle \frac{a}{b} = 1\Rightarrow a = b$

Similarly we can prove the same for $\displaystyle b$ and $\displaystyle c$ , thus, eventually proving that the condition for equality is $\displaystyle a=b=c$ .

Hence, equality occurs when $\displaystyle a = b = c$ for all positive integral values of $\displaystyle n$ .

$\displaystyle \Rightarrow C=3 \times \sqrt[3]{\frac{a^{3n}} {(2a)(2a)(2a)}}$

$\displaystyle \Rightarrow C=\frac{3}{2} \times a^{n-1}$

$\displaystyle a + b + c = 3 \ldots \{\text{ given condition} \}$

$\displaystyle \Rightarrow a = b = c = 1$ satisfies this condition as well as the inequality.

$\displaystyle \Rightarrow C=\frac{3}{2} \times (1)^{n-1}$

$\displaystyle \Rightarrow \boxed{C=1.5}$

$\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \geq C$

By Titu's Lemma/Cauchy Schwarz.

$\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \geq \frac{(a^{n/2}+b^{n/2}+c^{n/2})^2}{2(a+b+c)} \geq C$

Power mean inequality on the numerator

$\sqrt[\frac{n}{2}]{\frac{a^{n/2}+b^{n/2}+c^{n/2}}{3}} \geq \dfrac{a+b+c}{3}$

plugging in $a+b+c=3$

$a^{n/2}+b^{n/2}+c^{n/2} \geq 3$

$\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b} \geq \frac{(a^{n/2}+b^{n/2}+c^{n/2})^2}{2(a+b+c)}\geq \frac{(3)^2}{2(3)}=\frac{3}{2}=C$

please ignore my improper notation at the end :)

I did it in my head. Trying possibilities for a, b, and c ... then taking them to extrema. First tried a and b approaching zero, allowing c to approach 3 ... this rapidly grows without bound. So then dialed back until all three were equal, unity ... then of course simple to add 1/2 thrice.

I don't think I've seen anyone post a full solution with Chebyshev's inequality (surprising, considering it's on the page for it), so here's one:

WLOG, $a\ge b\ge c$ . Then $\frac{a^n}{b+c}\ge\frac{b^n}{c+a}\ge\frac{c^n}{a+b}$ which follows from $b+c\le c+a\le a+b$ .

Considering these two sequences, we have that by Chebyshev's inequality, $\left(\frac{\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}}{3}\right)\cdot\frac{(a+b)+(b+c)+(c+a)}{3}\ge \frac{(b+c)\cdot\frac{a^n}{b+c}+(c+a)\cdot\frac{b^n}{c+a}+(a+b)\cdot\frac{c^n}{a+b}}{3}$ .

This rearranges to $\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}\ge \frac{1}{2}(a^n+b^n+c^n)$ .

Then by the Power Mean Inequality, $\sqrt[n]{\frac{a^n+b^n+c^n}{3}}\ge \frac{a+b+c}{3}$ , which gives $a^n+b^n+c^n\ge 3$ .

From the previous inequality, we get $\frac{a^n}{b+c}+\frac{b^n}{c+a}+\frac{c^n}{a+b}\ge \boxed{\frac{3}{2}}$ , with equality iff $a=b=c=1$ .

It can also be done by chebyshev. Just prove that the expression with power n is less than or equal to the expression with power (n + 1). Then by nesbitt’s ineq, we have the result

Using Jensen's inequality, Let $\ f(x)=$ $\frac{x^n}{3-x}$ (Notice that $b + c = 3 - a$ ) Therefore, $\frac{f(a) + f(b) + f(c)}{3}$ $>=f$ ( $\frac{a+b+c}{3}$ )= $f (1)$ So $f(a) + f(b) + f(c) >=$ $\frac{3*1^n}{3-1}$ = $\frac{3*1^n}{2}$ . But since $n$ is an integer and the lowest value of $n$ to satisfy the inequality for all other choices of $n$ is one, replace $n$ as $1$ . $f(a) + f(b) + f(c) >=$ $\frac{3*1^1}{2}$ $=$ $\frac{3}{2}$ $=1.5$ = C . So our answer is $1.5$

f (x)= $\frac {x}{3-x}$ and then jenesen inequality will yield $3^n/2$ for minimum n=1 which gives 1.5