Question

Question: 4. Use Young's Inequality To Show That 3sin2θ≤2cos3θ+4sin23θ, For Θ∈(0,Π/2).

Answer 1

To prove the inequality 3sin2θ≤2cos3θ+4sin23θ, we can use Young's inequality to compare the terms.

Step-by-step explanation:

To prove the inequality 3sin2θ≤2cos3θ+4sin23θ, we can start by using Young's inequality. Young's inequality states that for any positive real numbers a, b, and p, we have ab ≤ (a^p)/p + (b^q)/q, where p and q are positive real numbers such that 1/p + 1/q = 1.

In this case, we can rewrite the inequality as
3sin2θ = 3(2sinθcosθ) = 6sinθcosθ ≤ (2cos3θ)^3/3 + (4sin23θ)^4/4.

By applying Young's inequality, we have
6sinθcosθ ≤ (2cos3θ)^3/3 + (4sin23θ)^4/4 <= 2(cos3θ)^2 + 4(sin23θ)^3.

We can simplify (cos3θ)^2 and (sin23θ)^3 to complete the proof.