can someone teach me how to prove algebraic problems? like this one,
show that if the sum of two positive numbers is 1, then the sum of their squares is ≥ 1/2 ?
$\displaystyle x+y=1\Longrightarrow y=1-x\,,\,\,x^2+y^2=(x+y)^2-2xy=1^2-2xy=1-2xy\geq 1-\frac{1}{2}=\frac{1}{2}$
Now, the last equality is true if $\displaystyle 2xy\leq \frac{1}{2}\Longleftrightarrow xy\leq \frac{1}{4}$ , so now just substitute $\displaystyle y=1-x$ here , get a
quadratic inequality in $\displaystyle x$ and check that iit is always true...
Hmmm...if you're a beginner the above may be a little hard (if you're in high school you shoud be able to do it, though), so perhaps there's
an easier way to do it WITHOUT the use of derivatives.
Also remember than in english, just as in spanish, german, russian, italian, etc., when you begin a sentence after a
period you must use capital letters.
Tonio
$\displaystyle x+y=1$ gives the relationship between x and y.
$\displaystyle x^2+y^2$ is the sum of the squares and we want it's minimum value.
Use $\displaystyle x+y=1$ to write the sum of squares in terms of x only.
$\displaystyle x+y=1\Rightarrow\ y=1-x,\;\;\;x<1,\;\;y<1$
$\displaystyle x^2+y^2=x^2+(1-x)^2=x^2+1-2x+x^2=2x^2-2x+1$
This is a U-shaped quadratic in x, valid for $\displaystyle 0<x<1$
The parabola is symmetrical about the minimum.
The minimum lies midway between the roots of the quadratic, even if the roots are complex,
which in this case they are as the graph lies entirely above the x-axis.
The roots of $\displaystyle ax^2+bx+c=0$ are
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The minimum lies midway between these, hence it lies at $\displaystyle x=-\frac{b}{2a}=-\frac{-2}{2(2)}=\frac{1}{2}$
$\displaystyle \left(\frac{1}{2}\right)^2+\left(1-\frac{1}{2}\right)^2=\frac{1}{4}+\frac{1}{4}$
Therefore $\displaystyle x^2+y^2\ \ge\ \frac{1}{2}$
A shortcut would be to find two equal values for $\displaystyle 2x^2-2x+1$
then find midway between them.
We can find f(x)=1
$\displaystyle 2x^2-2x+1=1\Rightarrow\ 2x^2-2x=0\Rightarrow\ 2x(x-1)=0$
$\displaystyle \Rightarrow\ x=1,\;\;\;x=0$
Midway between these is $\displaystyle x=\frac{1}{2}$
Therefore $\displaystyle f\left(\frac{1}{2}\right)$ gives the minimum.