AM-GM Practice 1

$10^x>0, \:\forall x\in\mathbb R$

Let $10^x=a$ . $\displaystyle 10^x-100^x=a(1-a)=-a^2+a$

There are 4 ways to continue:

1. Write it as $\displaystyle -a^2+a=-(a^2-a)=-((a-0.5)^2-0.25)$ $\displaystyle =-(a-0.5)^2+0.25\le \boxed{0.25}$

2. The vertex of the parabola is $a_v=-\frac{1}{2(-1)}=0.5\implies f(a_v)=0.5 (1-0.5)=0.5^2=\boxed{0.25}$

3. Using AM-GM (not exactly, since $1-a$ is not always non-negative. The inequality $ab\le \left(\frac{a+b}{2}\right)^2$ holds $\forall a,b\in\mathbb R$ ). $\displaystyle a(1-a)\stackrel{\text{AM-GM}}\le \left(\frac{a+(1-a)}{2} \right)^2=\boxed{0.25}$

4. Using derivatives.

In all the cases, the maximum is $\boxed{0.25}$ and it can be reached when $a=0.5\implies 10^x=0.5\implies x=-\log 2\approx -0.3$ . $\square$

Consideremos $10^x = k$ , então:

$10^x - 100^x = \\ 10^x - (10^x)^2 = \\ - k^2 + k =$

Ora, da expressão acima tiramos que ela possui um valor máximo; e o mesmo é dado por $Y_v = - \frac{\Delta}{4a}$

Segue que,

$Y_v = - \frac{\Delta}{4a} = \\\\ Y_v = - \frac{1^2 - 4 \cdot (- 1) \cdot 0}{4 \cdot - 1} = \\\\ Y_v = - \frac{1}{- 4} = \\\\ \boxed{Y_v = \frac{1}{4}}$

Let y = 10^x - 100^x, so 10^x = y + 100^x

Using AM-GM (y + 100^x)/2 >= 10^x(√y) But since 10^x = y + 100^x, dividing both sides by 10^x gives us 1/2 >= √y thus 1/4 = y is the maximum value.

10^{x}-100^{x}=10^{x}(1-10^{x})............................... {10^{x}(1-10^{x})}^(1/2) <= (10^{x}+1-10^{x})/2........................ 10^{x}(1-10^{x})<= 1/4