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 January 7th, 2018, 05:05 AM #1 Newbie   Joined: Jan 2018 From: Czechia Posts: 1 Thanks: 0 Prove inequality for every positive real number Hi, I have a problem which goes like this: Prove that for every positive real number t the following is valid: However, I have no idea how to solve it. Could anyone help me please?
 January 7th, 2018, 02:30 PM #2 Senior Member   Joined: Oct 2009 Posts: 863 Thanks: 328 Let's split it up in three subproblems: First, prove for all $t\geq 0$ that $$\frac{t^2 + 1}{t+1}\geq \sqrt{t}$$ Second, prove that for $0\leq t\leq 1$, that $$\frac{t^2 + 1}{t+1} - \sqrt{t} \leq 1-t$$ Finally, prove that for $t\geq 1$, that $$\frac{t^2 + 1}{t+1} - \sqrt{t} \leq t-1$$ Several ways of doing this. You could isolate the square root and then square both sides (but be careful with doing this, as you don't want to square negative numbers: -2<-1, but 4<1 is not true).
January 8th, 2018, 05:49 PM   #3
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Quote:
 Originally Posted by sleepysugar Hi, I have a problem which goes like this: Prove that for every positive real number t the following is valid: However, I have no idea how to solve it. Could anyone help me please?
A different way to split it up.

$\text {CASE I: } 0 < t < 1 \implies 0 < t^3 < t^2 < t < \sqrt{t} < 1.$

$0 < t < 1 \implies -\ 1 < t - 1 < 0 \implies 0 < |t - 1| < 1 \implies$

$|t - 1| = 1 - t \text { and } 0 < 1 - t < 1.$

$0 < t^2 < t \implies 0 < t^2 + 1 < t + 1 \implies 0 < \dfrac{t^2 + 1}{t + 1} < 1 \implies 0 < \dfrac{t^2 + 1}{t + 1} - t < 1 - t.$

$0 < t < \sqrt{t} \implies -\ \sqrt{t} < -\ t \implies \dfrac{t^2 + 1}{t + 1} - \sqrt{t} < \dfrac{t^2 + 1}{t + 1} - t < 1 - t \implies \dfrac{t^2 + 1}{t + 1} - \sqrt{t} \le |t - 1|.$

$0 < t^3 < t^2 < t < 1 \implies 0 < t^3 < 1 \text { and } 0 < 1 - t \implies$

$(1 - t) * 0 < (1 - t) * t^3 < (1 - t) * 1 \implies 0 < t^3 - t^4 < 1 - t \implies$

$t^3 + t < t^4 + 1 \implies t^3 + 2t^2 + t < t^4 + 2t^2 + 1 \implies$

$t(t^2 + 2t + 1) < t^4 + 2t^2 + 1 \implies t(t + 1)^2 < (t^2 + 1)^2 \implies$

$t < \dfrac{(t^2 + 1)^2}{(t + 1)^2} \implies \sqrt{t} < \dfrac{t^2 + 1}{t + 1} \implies 0 < \dfrac{t^2 + 1}{t + 1} - \sqrt{t}.$

$\therefore 0 < t < 1 \implies 0 \le \dfrac{t^2 + 1}{t + 1} - \sqrt{t} \le |t - 1|.$

$\text {CASE II: } 0 < 1 \le t.$

You can take it from here.

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