# For each value for a and b, being real numbers, a:2+b:2 >= ab

For each value for a and b, being real numbers, ${a}^{2}+{b}^{2}\ge ab$
Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing $2x+1$, etc...?
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Step 1
Try to write it as ${a}^{2}+{b}^{2}-ab=\frac{2{a}^{2}+2{b}^{2}-2ab}{2}=\frac{{a}^{2}+{b}^{2}+\left(a-b{\right)}^{2}}{2}$.
Step 2
Now all three terms are $\ge 0$, so you get ${a}^{2}+{b}^{2}\ge ab$.
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Kareem Mejia
Step 1
Alternative:
Either $\left(a×b\right)$ is negative, or it isn't.

Then, ${a}^{2}+{b}^{2}\ge 0>\left(a×b\right).$
Step 2

Then $2\left(a×b\right)\ge \left(a×b\right).$
Further, since $\left(a-b{\right)}^{2}\ge 0$ you have that
${a}^{2}+{b}^{2}\ge 2\left(a×b\right)\ge \left(a×b\right).$