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Prove that for all $x \in \mathbb{R}$ the inequality $2^x + 3^x - 4^x + 6^x - 9^x < 1$ holds.

Firstly, I rewrote the inequality as: $2^x + 3^x - (2^x)^2 + (2^x \cdot 3^x) - (3^x)^2 < 1$ and then tried to analyze a function for $x > 1$ and $x < 0$. Thank you for your responses.

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    have you tried substitution of $u = 2^x$ and $v = 3^x$? – 2017-01-06
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    Inequality is false for x=0. – 2017-01-06
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    I forgot to mention that there is an equality if and only if $x$=$0$ – 2017-01-06
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    Then the sign $\le$ seems more convenient. – 2017-01-06
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    Another one here: http://math.stackexchange.com/questions/563209/completion-of-the-square. – 2017-01-06
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    this one also appears in Putnam and Beyond – 2017-01-06

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