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pychu [463]
3 years ago
11

Prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
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astra-53 [7]

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Answer:

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Step-by-step explanation:

Using the midpoint formula

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