# Inequality and maximum

1. Dec 9, 2009

### Niles

Hi guys

Today my lecturer wrote on the blackboard

$$\max \left\{ {\alpha f(x) + (1 - \alpha )f(y)\,\,,\,\,\alpha g(x) + (1 - \alpha )g(y)} \right\}\,\,\,\, \le \,\,\,\alpha \max \left\{ {f(x)\,\,,\,\,g(x)} \right\} + (1 - \alpha )\max \left\{ {f(y)\,\,,\,\,g(y)} \right\},$$

where x, y are variables in all R, and alpha is a constant in [0;1]. I must admit, I cannot quite see why this inequality holds. Are there some rules about the maximum that is being used here?

2. Dec 9, 2009

### Lord Crc

Ok, starting with

$$\max \left\{ \alpha f(x) + (1 - \alpha )f(y),\alpha g(x) + (1 - \alpha )g(y) \right\}$$

Obviously, $$f(x) \le \max \left\{f(x), g(x)\right\} = x_{max}$$, and similar for $$g(x)$$, so we can safely use this to get something which is greater or equal:

$$\max \left\{ \alpha f(x) + (1 - \alpha )f(y),\alpha g(x) + (1 - \alpha )g(y) \right\} \le \max \left\{ \alpha x_{max} + (1 - \alpha )f(y),\alpha x_{max} + (1 - \alpha )g(y) \right\}$$

Again, $$f(y) \le \max \left\{f(y), g(y)\right\} = y_{max}$$, and similar for $$g(y)$$, and using that we get

$$\max \left\{ \alpha x_{max} + (1 - \alpha )f(y),\alpha x_{max} + (1 - \alpha )g(y) \right\} \le \max \left\{ \alpha x_{max} + (1 - \alpha )y_{max},\alpha x_{max} + (1 - \alpha )y_{max} \right\}$$

Now, for $$a \ge 0$$ we have that $$\max \left \{ ab, ac \right \} = a \max \left \{b, c \right \}$$ and $$\max \left \{ a + b, a + c \right \} = a + \max \left \{b, c \right \}$$. Assuming $$\alpha \in [0,1]$$, we can use this to get

$$\max \left\{ \alpha x_{max} + (1 - \alpha )y_{max},\alpha x_{max} + (1 - \alpha )y_{max} \right\} = \alpha x_{max} + (1 - \alpha )\max \left\{y_{max}, y_{max} \right\} = \alpha x_{max} + (1 - \alpha )y_{max}$$.

Inserting the expressions for $$x_{max}$$ and $$y_{max}$$ gets you to the right hand side you had, which thus is equal or greater than the left hand side.

3. Dec 10, 2009

### Niles

Thank you. It is very kind of you to help.