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How can we prove the following inequality

$$\left (3x+\frac{4}{x+1}+\frac{8}{\sqrt{2(y^2+1)}}\right )\left (3y+\frac{4}{y+1}+\frac{8}{\sqrt{2(x^2+1)}}\right )\geq 81,\ \forall x,y\geq 0$$

?

I have proved that: $3x+\frac{4}{x+1}+\frac{8}{\sqrt{2(x^2+1)}}\geq 9, \forall x\geq 0$, with equality only when $x=1$.

Bogdan
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    Are you sure about your equations? You have $x$ in your proved equation below the square root, but your original equation reads $y$ there. – EarthwormA3aan Mar 04 '16 at 09:39
  • You proved that the arithmetic mean of (3x + 4/(x+1)) and 8/(...) is at least 9/2. If you prove that their geometric mean is at least 9/2, it solves the problem. – zyx Mar 04 '16 at 09:42
  • Continuation of http://math.stackexchange.com/questions/1682088/weird-inequality-that-seems-to-be-true ? – Macavity Mar 04 '16 at 09:42
  • The LHS is a symmetric function wrt $x$ and $y$, hence we just need to prove that the minima of the LHS occur along the $x=y$ line. – Jack D'Aurizio Mar 04 '16 at 09:42
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    @JackD'Aurizio That's a fallacy. For a counter eg check his linked older post. – Macavity Mar 04 '16 at 10:05
  • @Macavity: what fallacy? I am just saying that it is enough to study the behaviour of that function along the lines $x+y=k$, but obviously a symmetric function may have absolute minima at $(x_0,y_0)$ and $(y_0,x_0)$ with $x_0\neq y_0$. I do not know if it is the case for our symmetric function. – Jack D'Aurizio Mar 04 '16 at 13:16
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    See here: http://math.stackexchange.com/questions/1596353/ – Michael Rozenberg Mar 04 '16 at 19:34

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