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...?

Yaretzi Mcconnell
2022-11-15
Answered

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...?

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...?

You can still ask an expert for help

Calvin Maddox

Answered 2022-11-16
Author has **15** answers

Step 1

Try to write it as ${a}^{2}+{b}^{2}-ab=\frac{2{a}^{2}+2{b}^{2}-2ab}{2}=\frac{{a}^{2}+{b}^{2}+(a-b{)}^{2}}{2}$.

Step 2

Now all three terms are $\ge 0$, so you get ${a}^{2}+{b}^{2}\ge ab$.

Try to write it as ${a}^{2}+{b}^{2}-ab=\frac{2{a}^{2}+2{b}^{2}-2ab}{2}=\frac{{a}^{2}+{b}^{2}+(a-b{)}^{2}}{2}$.

Step 2

Now all three terms are $\ge 0$, so you get ${a}^{2}+{b}^{2}\ge ab$.

Kareem Mejia

Answered 2022-11-17
Author has **9** answers

Step 1

Alternative:

Either $(a\times b)$ is negative, or it isn't.

$\underset{\_}{\text{Case 1:}\text{}(a\times b)\text{}\text{is negative}}$

Then, ${a}^{2}+{b}^{2}\ge 0>(a\times b).$

Step 2

$\underset{\_}{\text{Case 2:}\text{}(a\times b)\text{}\text{is not negative}}$

Then $2(a\times b)\ge (a\times b).$

Further, since $(a-b{)}^{2}\ge 0$ you have that

${a}^{2}+{b}^{2}\ge 2(a\times b)\ge (a\times b).$

Alternative:

Either $(a\times b)$ is negative, or it isn't.

$\underset{\_}{\text{Case 1:}\text{}(a\times b)\text{}\text{is negative}}$

Then, ${a}^{2}+{b}^{2}\ge 0>(a\times b).$

Step 2

$\underset{\_}{\text{Case 2:}\text{}(a\times b)\text{}\text{is not negative}}$

Then $2(a\times b)\ge (a\times b).$

Further, since $(a-b{)}^{2}\ge 0$ you have that

${a}^{2}+{b}^{2}\ge 2(a\times b)\ge (a\times b).$

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