How to prove this inequality in interval: (0,π2) 2cos⁡x+sec2x−3≥0

Question

How to prove this inequality in interval: (0,π2)  2cos⁡x+sec2x−3≥0

Thomas Nickerson · Accepted Answer

Restrict the domain to some meaningful interval such as

x \in (- \frac{π}{2}, \frac{π}{2})

so that

\cos x > 0

. In that case, using the AM-GM inequality you have:

L H S = \cos x + \cos x + \sec^{2} x - 3 \geq 3 \sqrt[3]{\cos x \cdot \cos x \cdot \sec^{2} x} - 3 = 3 - 3 = 0 = R H S

Papilys3q · Accepted Answer

We have  2cos⁡x+sec2x−3=2(cos⁡x−1)+(1cos2x−1)=(2cos⁡x+1)(cos⁡x−1)2cos2x  Thefore the inequality does not hold.

Vasquez · Accepted Answer

By Fermat&#039;s theorem, we examine zeroes of derivative of the function f(x)=2cos⁡x+sec2⁡x−3 to find the extrema, that is, to solve the following: f′(x)=2tan⁡xsec2⁡x−2sin⁡x=0 The solution is x={kπ|k∈Z}. When x=π, we get 2cos⁡(x)+sec2⁡(x)−3=−4 Thus 2cos⁡x+sec2⁡x−3≥0 doesn&#039;t hold. Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when x∈(0,π2), it is trivial that f′(x)≥0, thus f(x) monotonously increases on (0,π2). Hence f(x)≥f(0)=0 for all x∈(0,π2) and we finished the proof.

How to prove this inequality in interval: (0,\frac{\pi}{2}) \ \

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