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Math Help - Prove an inequality

  1. #1
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    Prove an inequality

    Hello, could someone help with proving this inequality:

    a^2+b^2+1 \ge ab + a + b

    Thanks in advance
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  2. #2
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    One way is to consider a line b = k * a for some number k and to prove that the inequality holds for all points (a, b) on this line. (For completeness, one should consider the line a = 0, but the inequality is symmetric w.r.t. a and b, so this is similar to k = 0.)

    The substitution b = ka gives rise a quadratic inequality on a where coefficients are quadratic polynomials of k. By considering the leading coefficient and the discriminant, one can show that the inequality holds.
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  3. #3
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    Quote Originally Posted by mat1990 View Post
    Hello, could someone help with proving this inequality:

    a^2+b^2+1 \ge ab + a + b

    Thanks in advance
    You could also utilise the Arithmetic Mean -- Geometric Mean Inequality

    \displaystyle\frac{a+b}{2}\ge\sqrt{ab}

    Therefore

    \displaystyle\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}

    \displaystyle\frac{a^2+1}{2}\ge\sqrt{a^2}

    \displaystyle\frac{b^2+1}{2}\ge\sqrt{b^2}

    Hence, in summing these

    a^2+b^2+1\ge\ |ab|+|a|+|b|\Rightarrow\ a^2+b^2+1\ge\ ab+a+b
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  4. #4
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    Quote Originally Posted by mat1990 View Post
    Hello, could someone help with proving this inequality:

    a^2+b^2+1 \ge ab + a + b

    Thanks in advance
    A more direct way to prove the inequality is...

    (a-b)^2+(a-1)^2+(b-1)^2\ge\ 0

    \Rightarrow\ a^2-2ab+b^2+a^2-2a+1+b^2-2b+1\ge\ 0

    2\left(a^2+b^2+1\right)\ge\ 2(ab+a+b)
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