Working with proofs help? I'm trying to study for my midterm and doing some random practise questions to work with proofs. However I'm stuck on, as the only way I know how to prove it is through plugging in numbers, however as I understand that is not solid proof.
Here it is:
If $x$ is a real number and $x > 0$. Prove that $$x+\frac{9}{x} \geq 6$$.
Any help would be greatly appreciated.
 A: We assume $x > 0$, start by multiplying through by $x$. (This is legitimate because in we know $x \gt 0,$ and multiplying an inequality by a positive, real number $x$, will not change the direction of the inequality.) 
Doing so, and then gathering all terms to the left of the inequality gives us:  $$x + 9/x \geq 6 \iff x^2 + 9 \geq 6x \iff x^2 - 6x + 9 \geq 0\iff (x - 3)^2 \geq 0,$$ which is true for all real values $\,x > 0\,$ (and those are the only real values of concern) simply by virtue that the square of a number is greater than or equal to $0$.
$$\textrm{Hence, if}\;\,x\in \mathbb{R}, x\gt 0, \;\;\text{ then}\;\; x + \dfrac 9x \geq 6$$
A: Transform it to quadratic inequality: $x^2-6x+9\geq 0$
Solve it and you will get solution - all real numbers.
Or simply write it as square which always is equal or bigger than $0$: $(x-3)^2\geq 0$
A: hint:$ax^2+bx+c \ge0 $ when $b^2-4ac\le o$ and $a\ge 0$ $$ x+\frac{9}{x}\ge 6$$ then $$x^2-6x+9\ge0$$
A: Let $y=x+\frac {a^2}x$ where real $a>0$ 
$\implies x^2-xy+a^2=0 $
This is a quadratic equation in $x$ and if $x$ is real the discriminant $y^2-4\cdot1\cdot a^2$ must be $\ge0\implies y^2\ge 4a^2$ 
$\implies $ either $y\ge 2a$ or $y\le -2a$
If $x>0,y>0\implies y\ge2a$
We can also use calculus here 
$\frac{dy}{dx}=1-\frac{a^2}{x^2}$
For the extreme values of $y,\frac{dy}{dx}=0\implies x=\pm a$
But as $x>0,x$ can not be $-a$
Now,  $\frac{d^2y}{dx^2}=\frac{2a^2}{x^3}>0$ as $x>0$
So, at $x=a,y$ will attain its minimum value  $a+\frac{a^2}{a}=2a$
