If a and B are real numbers, prove that a ^ 2 + B ^ 2 is greater than or equal to 1 / 2 (a + b) ^ 2

If a and B are real numbers, prove that a ^ 2 + B ^ 2 is greater than or equal to 1 / 2 (a + b) ^ 2

2a²+2b²-（a+b）²
=a²+b²-2ab
=(a-b)²≥0
∴2a²+2b²≥（a+b）²
That is a & sup2; + B & sup2; ≥ 1 / 2 (a + b) & sup2;

Given that a and B are positive real numbers and a + B = 1, it is proved that (a + 1 / 2) ^ 2 + (B + 1 / 2) ^ 2 is less than or equal to 2

Square of (√ (a + 1 / 2) + √ (B + 1 / 2)) = a + B + 1 + 2 √ (a + 1 / 2) * (B + 1 / 2)
=2+2√（ab+1/2(a+b)+1/4),
Since a + B = 1, ab

Given that a and B are positive real numbers, a + B = 1, prove (a + 1 / a) * (B + 1 / b) > = 25 / 4

Because a + B = 1, the original formula = (2a + B / a) * (2B + A / b) = (2 + B / a) * (2 + A / b) = 4 + 2 (A / B + B / a) + 1 = 5 + 2 (A / B + B / a) > = 5 + 2 * 2 = 9 (if and only if a = b = 1 / 2, take the equal sign), that is, the original formula > = 9. I don't know why the original formula should prove > = 25 / 4? Obviously 9 > 25 / 4, so (a + 1 / a) * (B + 1 / b) > = 25 / 4