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Definite Integration: Constant Minimum Value

JEE Maths question with a full step-by-step solution.

Question
If 3x2+1x4+x2≥max⁡f(x)\dfrac{3x^2+1}{\sqrt{x^4+x^2}} \ge \max f(x), where f(x)f(x) is constant for 0<x≤20<x\le\sqrt2, then the minimum value of
∫023x2+1x4+x2 dx\int_{0}^{\sqrt2}\frac{3x^2+1}{\sqrt{x^4+x^2}}\,dx
 is
Solution
Answer: 4
Step 1:
x4+x2=x2(x2+1)=xx2+1(x>0)\sqrt{x^4+x^2} = \sqrt{x^2\left(x^2+1\right)} = x\sqrt{x^2+1} \qquad (x>0)
 Step 2:
3x2+1=2x2+(x2+1)3x^2+1 = 2x^2+\left(x^2+1\right)
3x2+1xx2+1=2x2xx2+1+x2+1xx2+1=2xx2+1+x2+1x\frac{3x^2+1}{x\sqrt{x^2+1}} = \frac{2x^2}{x\sqrt{x^2+1}}+\frac{x^2+1}{x\sqrt{x^2+1}} = \frac{2x}{\sqrt{x^2+1}}+\frac{\sqrt{x^2+1}}{x}
 Step 3: Both terms are positive for x>0x>0, and their product is
2xx2+1⋅x2+1x=2\frac{2x}{\sqrt{x^2+1}}\cdot\frac{\sqrt{x^2+1}}{x} = 2
 a constant, so by AM-GM the sum is at least 222\sqrt2:
3x2+1x4+x2≥22for every x>0\frac{3x^2+1}{\sqrt{x^4+x^2}} \ge 2\sqrt2 \qquad\text{for every }x>0
 Step 4: Equality needs the two terms equal:
2xx2+1=x2+1x  ⟹  2x2=x2+1  ⟹  x=1\frac{2x}{\sqrt{x^2+1}} = \frac{\sqrt{x^2+1}}{x} \;\Longrightarrow\; 2x^2 = x^2+1 \;\Longrightarrow\; x = 1
 and 1∈(0,2]1 \in \left(0,\sqrt2\right], so 222\sqrt2 is the greatest constant lower bound. Step 5:
∫023x2+1x4+x2 dx ≥ ∫0222 dx=22(2−0)=4\int_{0}^{\sqrt2}\frac{3x^2+1}{\sqrt{x^4+x^2}}\,dx \ \ge\ \int_{0}^{\sqrt2}2\sqrt2\,dx = 2\sqrt2\left(\sqrt2-0\right) = 4
 Answer: 4.004.00
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