Prove that a+b/2≥√ab
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Start by squaring both sides got remove the sqrt
[(a+b)/2]^2 ≥ ab
(a^2 +2ab +b^2 )/4≥ ab
Multiply both sides by 4
a^2 +2ab +b^2 ≥ 4ab
Subtract 2ab from both sides
a^2 + b^2 ≥ 2ab
Subtract 2ab from both sides
a^2 - 2ab + b^2 ≥ 0
Use quadratic equation to solve where a = 1 and b = -2b
2b ± sqrt[(-2b)^2 - (4x1xb^2)]/(2x1)
2b ± sqrt[4b^2 - 4b^2]/2
(2b ± 0)/2
a = b
Substituting for b in the equation a^2 + b^2 ≥ 2ab
2a^2 ≥ 2a^2

121.1

Step-by-step explanation:

By law of sines, (sin <I)/i = (sin <J)/j. Plugging in the values we know, we get sin 72deg / 206 = sin 34deg / j. When we solve for k, we get that it is about 121.122. Rounded to the nearest tenth, we get that j is 121.1.