Metodo de Jensen, localização relativa e metodo das secantes

Recall that a convex function is a function whose secant line lies above the graph. Call this function $f(x)$. Then, by the definition, if we draw a secant line that passes through the points $(x_1 , f(x_1))$, and $(x_2 , f(x_2))$, note that if we take a weighted average of $x_1$ and $x_2$, which equals $\lambda x_{1} + (1-\lambda)x_{2}$. The corresponding $y$ value on the graph of $f(x)$ with the weighted average is $f(\text{average}) = f(\lambda x_1 + (1-\lambda)x_2 )$. Also, note that the corresponding $y$ value of the secant line is the weighted average of $f(x_1 )$ and $f(x_2 )$ with the same weights, or $\lambda f(x_1 ) + (1-\lambda )f(x_2 )$. This means that $$\lambda f(x_1 ) + (1-\lambda )f(x_2 ) \geq f(\lambda x_1 + (1-\lambda)x_2 )$$ I will refer to this as "convexity inequality". Note that the inequality is reversed when $f$ is concave.

Jensen's weighted inequality is: For non-negative real numbers $\lambda_{1}, \lambda_{2}, \cdots , \lambda_{n}$, and a convex function $f(x)$, and real numbers $x_1 , x_2 , \cdots , x_n$ $$\lambda_{1}f(x_{1}) + \lambda_{2}f(x_{2}) + \cdots + \lambda_{n}f(x_{n}) \geq f(\lambda_{1}x_{1} + \lambda_{2}x_{2} + \cdots +\lambda_{n}x_{n})$$c The proof proceeds from induction. The base case where $n = 2$ is an immediate consequence of the convexity inequality. Now, suppose the inequality is true for a positive real $n$. Then it suffices to show that the inequality is true for $n + 1$ and we are done. We write $$f\left(\sum_{i = 1}^{n+1}\lambda_{i}x_{i}\right) = f\left(\lambda_{1}x_{1} + (1-\lambda_{1})\sum_{i=2}^{n+1}\frac{\lambda_{i}}{1-\lambda_{1}}x_{i}\right)$$ Applying the convexity inequality, $$f\left(\lambda_{1}x_{1} + (1-\lambda_{1})\sum_{i=2}^{n+1}\frac{\lambda_{i}}{1-\lambda_{1}}x_{i}\right) \leq \lambda_{1}f(x_{1}) + (1-\lambda_{1})f\left( \sum_{i = 2}^{n+1}\frac{\lambda_{i}}{1 - \lambda_{1}}x_{i}\right)$$ And since $\sum_{i =2}^{n+1} \frac{\lambda_{i}}{1-\lambda_{1}} = 1$, and applying the assumed information, we have the desired inequality, and we are done.

Some problems:

Problem 1: Find the maximum value of $\sqrt{x+1} + \sqrt{y + 1} + \sqrt{z+1}$ if $x + y + z = 3$, and $x, y, z >0$

Solution 1: If $f(x) = \sqrt{x + 1}$, $f(x)$ is concave on the interval, so by Jensen's inequality, $$\frac{1}{3}f(x) + \frac{1}{3}f(y) + \frac{1}{3}f(z) \leq f(\frac{x+y+z}{3}) = \sqrt{2}$$ with equality when $x=y=z = 1$, so the maximum value is $\boxed{3\sqrt{2}}$

Problem 2 (Japan MO Finals #3): If $a + b + c = 1$, and $a,b,c >0$ prove that $$a\sqrt[3]{1 - b + c} + b\sqrt[3]{1 - c + a} + c\sqrt[3]{1 - a + b} \leq 1$$

Solution 2: Let $f(x) = \sqrt[3]{x}$. Then by Jensen's, since $a + b + c = 1$, and $f(x)$ is concave on the interval, we have $$af(1-b+c) + bf(1-c+a) + cf(1 - a + b) \leq$$ $$f(a - ab + ac + b - bc + ab + c - ac + bc) = f(a + b + c) = f(1) = 1$$ which is the desired inequality. For the relevant AoPS form on this problem, see http://artofproblemsolving.com/community/c6h31799p197379

Problem 3 (IMO 2001 #2): Prove that, for $a, b, c > 0$, $$\frac{a}{\sqrt{a^{2} + 8bc}} + \frac{b}{\sqrt{b^{2} + 8ac}} + \frac{c}{\sqrt{c^{2} + 8ab}} \geq 1$$

Solution 3: Since the inequality is homogeneous, let $a + b + c = 1$. Then, let $f(x) = \frac{1}{\sqrt{x}}$. Note that on the interval, $f(x)$ is convex, so from Jensen's, we have $$af(a^{2} + 8bc) + bf(b^{2} + 8ac) + cf(c^{2} + 8ab) \geq f(a^{3} + b^{3} + c^{3} + 24abc)$$ Now it suffices to show that the RHS is greater than or equal to $1$. In other words, we want to prove $$\frac{1}{\sqrt{a^{3}+b^{3} + c^{3} + 24abc}} \geq 1$$ which is equivalent to proving $$a^{3} + b^{3} + c^{3} + 24abc \leq 1 = (a+b+c)^{3}$$ After subtracting $a^{3} +b^{3} + c^{3} + 6abc$ from both sides and dividing both sides by $3$, we are done by $AM-GM$. For the relevant form on AoPS, see http://artofproblemsolving.com/community/c3222_imo