# Hàm lồi và cực trị của biểu thức dạng xtanA+ytanB+ztanC

Posted by Nguyen Song Minh

Edited by Nguyen Trung Tuan

In this post $I = (a,b)\subset \mathbb{R}$.
Theorem. If $f: I\to\mathbb{R}$ such that $f''(x) > 0\forall x\in I$ then $f(c)\geq (c - d)f'(d) + f(d)$. Equal hold iff $c = d$. Where $c,d\in I$.

Proof. Cases follows shall occur
a)If $c = d$, we have equal,
b)If $c > d$ then $\dfrac {f(c) - f(d)}{c - d} = f'(m)$ for some $m\in (d,c)$, since $f''(x) > 0\forall x\in I$ hence $f'(m) > f'(d)$ and we have done!
c)If $c < d$ similary!

Corollary $1$. Let $f_{i}: I\to\mathbb{R}(i = 1,2,...,n)$  such that $f''_{i}(x) > 0\forall i = 1,2,...,n\forall x\in I$. Let $x_{1},x_{2},...,x_{n};y_{1},y_{2},...,y_{n}\in I$ such that $x_{1} + x_{2} + ... + x_{n} = y_{1} + y_{2} + ... + y_{n}$ and $f'_{i}(y_{i}) = f'_{j}(y_{j})\forall i,j = 1,2,...,n$. Then we have $f_{1}(x_{1}) + f_{2}(x_{2}) + ... + f_{n}(x_{n})\geq f_{1}(y_{1}) + f_{2}(y_{2}) + ... + f_{n}(y_{n})$ and equal holds iff $x_{i} = y_{i}\forall i = 1,2,...,n$.

Proof. Because $f_{i}(x_{i})\geq (x_{i} - y_{i})f'_{i}(y_{i}) + f_{i}(y_{i})\forall i = 1,2,...,n$ therefore we have done.

Corollary $2$. Let acute triangles $ABC$ and $MNP$ , then we have $\cos^{2}{M}\tan{A} + \cos^{2}{N}\tan{B} + \cos^{2}{P}\tan{C}\geq\\ \frac {1}{2}(\sin{2M} + \sin{2N} + \sin{2P})$.

Proof. Using Corollary $1$ with three functions $f_{M}(x) = \cos^{2}{M}\tan{x},f_{N}(x) = \cos^{2}{N}\tan{x},\\ f_{P}(x) = \cos^{2}{P}\tan{x},I = (0,\frac {\pi}{2})$ and $x_{1} = A,x_{2} = B,x_{3} = C;y_{1} = M,y_{2} = N,y_{3} = P$.

Now I will use Corollary $2$ to solution problem follows

Problem(VMEO). Find minimum of $\tan{A} + 2\tan{B} + 5\tan{C}$ where $A,B,C$ are three angles of an acute triangle $ABC$.

Solution. By Corollary $2$ we need find the acute triangle $MNP$ such that $\dfrac {\cos^{2}{M}}{1} = \dfrac {\cos^{2}{N}}{2} = \dfrac {\cos^{2}{P}}{5}$. This is easy job, in fact, denote common value of fractions above is $k$ then by $\cos^{2}{M} + \cos^{2}{N} + \cos^{2}{P} + 2\cos{M}\cos{N}\cos{P} = 1$ we have $k = \frac {1}{10}$.

Therefore $\tan{A} + 2\tan{B} + 5\tan{C}\geq \\ 10.\frac {1}{2}(\sin{2M} + \sin{2N} + \sin{2P}) = 12$.

P/S: We can find minimum of $x\tan{A} + y\tan{B} + z\tan{C}$ by using Corollary $2$.

## One thought on “Hàm lồi và cực trị của biểu thức dạng xtanA+ytanB+ztanC”

1. Jon Dattorro says:

Convex Optimization & Euclidean Distance Geometry

http://meboo.convexoptimization.com