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Math Help - inequality

  1. #1
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    inequality [SOLVED]

    Hello
    is there any elegant way to prove the following,
    \sqrt{2a+1}+\sqrt{2b+1}\leq a+b+2,(a,b)\in \mathbb{R}^{+}
    Thanks
    Last edited by Raoh; November 25th 2010 at 05:51 PM.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    Hello
    is there any elegant way to prove the following,
    \sqrt{2a+1}+\sqrt{2b+1}\leq a+b+2,(a,b)\in \mathbb{R}^{+}
    Thanks
    Dear Raoh,

    Note that, (a+1)^2=a^2+2a+1\geq{2a+1}~whenever~a\in\Re^{+}

    \Rightarrow{a+1\geq\sqrt{2a+1}}~whenever~a\in\Re^+

    Similarly, \Rightarrow{b+1\geq\sqrt{2b+1}}~whenever~a\in\Re^+

    Therefore, \Rightarrow{a+1+b+1\geq\sqrt{2a+1}+\sqrt{2b+1}}~wh  enever~a,b\in\Re^+

    a+b+2\geq\sqrt{2a+1}+\sqrt{2b+1}~whenever~a,b\in\R  e^+

    Hope this helps.
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  3. #3
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    i prefer this approach,
    \sqrt{2a+1}+\sqrt{2b+1}\leq  \sqrt{2a+1+a^2}+\sqrt{2b+1+b^2}=a+b+2
    Your solution was awesome.
    Thank you.
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