Exercise
Prove the inequality:
a b ≤ a + b 2 \sqrt{ab}\leq \frac{a+b}{2} a b ≤ 2 a + b
Note: this inequality is the n=2 case in Inequality of arithmetic and geometric means.
( a 1 ⋅ a 2 ⋅ . . . ⋅ a n ) 1 n ≤ a 1 + a 2 + . . . + a n n {(a_1\cdot a_2 \cdot ...\cdot a_n)}^{\frac{1}{n}}\leq\frac{a_1+a_2+...+a_n}{n} ( a 1 ⋅ a 2 ⋅ . . . ⋅ a n ) n 1 ≤ n a 1 + a 2 + . . . + a n
Proof
We use short multiplication furmula
( a − b ) 2 = a 2 − 2 a b + b 2 {(a-b)}^2=a^2-2ab+b^2 ( a − b ) 2 = a 2 − 2 a b + b 2
The following sentence is true for all a and b:
0 ≤ ( a − b ) 2 0\leq {(a-b)}^2 0 ≤ ( a − b ) 2
Since every squared expression is positive.
Open the brackets:
0 ≤ a 2 − 2 a b + b 2 0\leq a^2-2ab+b^2 0 ≤ a 2 − 2 a b + b 2
We express the middle expression as a sum and we get:
0 ≤ a 2 − 4 a b + 2 a b + b 2 0\leq a^2-4ab+2ab+b^2 0 ≤ a 2 − 4 a b + 2 a b + b 2
Rearrange:
0 ≤ a 2 + 2 a b + b 2 − 4 a b 0\leq a^2+2ab+b^2-4ab 0 ≤ a 2 + 2 a b + b 2 − 4 a b
We use short multiplication furmula
0 ≤ ( a + b ) 2 − 4 a b 0\leq {(a+b)}^2-4ab 0 ≤ ( a + b ) 2 − 4 a b
4 a b ≤ ( a + b ) 2 4ab\leq {(a+b)}^2 4 a b ≤ ( a + b ) 2
a b ≤ ( a + b ) 2 4 ab\leq \frac{{(a+b)}^2}{4} a b ≤ 4 ( a + b ) 2
a b ≤ a + b 2 \sqrt{ab} \leq \frac{a+b}{2} a b ≤ 2 a + b
And we got the desired result.
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