Math IOQM

 Good tutorial here . Proof using Cauchy Shwarz . Example:

  First focus on a^5 + b^5. WLOG a >= b => a^2 >= b^2 => a^3 >= b^3 By rearrangement inequality : a^2.a^3 + b^2.b^3 >= a^2.b^3 + b^2.a^3 => a^5 + b^5 >= a^2.b^3 + b^2.a^3 Add ab to both sides to get the first denominator: a^5+ b^5 + ab >= a^2.b^3 + b^2.a^3 + ab = ab(1 + a^2.b + a.b^2) Cross multiplication: 1/(1 + a^2.b + a.b^2) >= ab/(a^5+ b^5 + ab) Now we can see that RHS is the first term of the original question expression. Now if we put abc = 1 in LHS: then LHS = abc/(abc + a^2.b + a.b^2) Divide numerator/denominator by ab: LHS = c/(c + a + b) So: c/(c + a + b) >= ab/(a^5+ b^5 + ab) Similarly: b/(c + a + b) >= ac/(a^5+ c^5 + ac) a/(c + a + b) >= bc/(b^5+ c^5 + bc) Add all: LHS = 1 RHS is the original question expression. => 1 >= original expression. Hence proved. In the above question we used Cauchy Shwarz Inequality . We will use Titu's Lemma here. Above to prove that x^4 + 2x >= 3x^2 We can use A.M. G.M. as well. Apply AM GM on...

  Idea in plain words Line up two lists of numbers that both go up, like heights: a 1 ≤ a 2 ≤ ⋯ ≤ a n a_1 \le a_2 \le \dots \le a_n backpack weights: b 1 ≤ b 2 ≤ ⋯ ≤ b n b_1 \le b_2 \le \dots \le b_n Now pair the first height with the first weight, the second with the second, etc. Product of averages ( a 1 + ⋯ + a n n ) ( b 1 + ⋯ + b n n ) \displaystyle \left(\frac{a_1+\cdots+a_n}{n}\right)\left(\frac{b_1+\cdots+b_n}{n}\right) is like saying: “Pretend every student has the average height and carries the average backpack. What would the ‘typical’ height×weight be?” Average of products a 1 b 1 + a 2 b 2 + ⋯ + a n b n n \displaystyle \frac{a_1b_1+a_2b_2+\cdots+a_nb_n}{n} is what you actually get when the shorter students carry the lighter backpacks and the taller students carry the heavier ones. Chebyshev’s inequality says: when both lists rise together (small with small, big with big), average of products       ≥       product of a...

Power mean inequality: Power Mean  For real numbers x 1 , x 2 , … , x n ≥ 0 x_1, x_2, \ldots, x_n \ge 0 and real p ≠ 0 p \ne 0 , the power mean of order p p is: M p = ( x 1 p + x 2 p + ⋯ + x n p n ) 1 / p M_p = \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n} \right)^{1/p} Power Mean Inequality If p < q p < q , then: ( x 1 p + x 2 p + ⋯ + x n p n ) 1 / p ≤ ( x 1 q + x 2 q + ⋯ + x n q n ) 1 / q \left( \frac{x_1^p + x_2^p + \cdots + x_n^p}{n} \right)^{1/p} \le \left( \frac{x_1^q + x_2^q + \cdots + x_n^q}{n} \right)^{1/q} with equality if and only if x 1 = x 2 = ⋯ = x n x_1 = x_2 = \cdots = x_n . Example (n = 3, p = 2, q = 3) Let’s say: x 1 , x 2 , x 3 ≥ 0 x_1, x_2, x_3 \ge 0 Then the power mean inequality says: ( x 1 2 + x 2 2 + x 3 2 3 ) 1 / 2 ≤ ( x 1 3 + x 2 3 + x 3 3 3 ) 1 / 3 \left( \frac{x_1^2 + x_2^2 + x_3^2}{3} \right)^{1/2} \le \left( \frac{x_1^3 + x_2^3 + x_3^3}{3} \right)^{1/3} Ex1) Prove that a 4 + b 4 + c 4 ≥ a b c ( a + b + c ) a^4 + b^4 + c^4...