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How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$ see last answer why did he add 16lnx but lnx doesnt exist

Saurav Maheshkar
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Will
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  • but isn't the topic is old enough? also, my point is less than 50 – Will Mar 29 '22 at 08:01
  • could u comment it for me i dont have enough point to comment there – Will Mar 29 '22 at 08:03
  • He added $\ln(x)$ since he then considers $f(a)+f(b)+f(c) $, where the extra term gives $\ln(a) + \ln(b) + \ln(c) = \ln (a b c) = \ln (1) = 0$, since the constraint in the linked question is $a b c = 1$. So in fact he added zero. – Andreas Mar 29 '22 at 10:14
  • owh thanks to make it clear – Will Mar 29 '22 at 10:21

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This is a well-known trick for some inequalities of the form $f(a) + f(b) + f(c) \ge 0$ under the condition $a, b, c > 0; abc = 1$, assuming that the equality case is $a = b = c = 1$.

The idea is: If we can find a real constant $m$ such that $F(x) := f(x) + m \ln x \ge 0$ for all $x > 0$, then we have $0 \le F(a) + F(b) + F(c) = f(a) + f(b) + f(c) + m \ln (abc) = f(a) + f(b) + f(c)$. The desired inequality is proved. (If we can not find such a $m$, this trick fails.)

How to find $m$? Since $F(1) = 0$, if $F(x) \ge 0$ for all $x > 0$, then $x = 1$ is the minimum point of $F(x)$ on $x > 0$ which results in $F'(1) = 0$. From $F'(1) = 0$, we have $f'(1) + m = 0$ and $m = - f'(1)$. It remains to prove that $f(x) - f'(1)\ln x \ge 0$ for all $x > 0$.

For the problem you refer to, the inequality is written as $f(a) + f(b) + f(c) \ge 0$ with $f(x) = x^2 + \frac{8}{x} + 1 - 10x$.

We have $f'(1) = -16$. We need to prove that $f(x) + 16\ln x \ge 0$ for all $x > 0$. Fortunately, it is true.

There are lots of examples in AoPS and MSE, etc. Here are some.

Problem 1 (jokehim@AoPS): Let $a, b, c > 0$ with $abc = 1$. Prove that $$\sum_{\mathrm{cyc}} \frac{a^3}{\sqrt{1+a^4}} \ge \sum_{\mathrm{cyc}} \frac{\sqrt2}{a^2+1}.$$

It suffices to prove that, for all $x > 0$, $$\frac{x^3}{\sqrt{1+x^4}} - \frac{\sqrt2}{x^2+1} - \frac{3}{\sqrt2}\ln x \ge 0.$$

Problem 2: Let $n \ge 3$ be a positive integer. Let $a, b$ be positive real numbers such that $a^{n+1} + b^{n+1} = 2$. Prove that $a^n+b^n \ge a^{n-1} + b^{n-1}$.

It suffices to prov that, for all $x \in [0, 2]$, $$x^n - x^{n - 1} \ge \frac{1}{n+1}(x^{n+1} - 1).$$

Note: The idea is similar, i.e. to find a real constant $m$ such that $x^n - x^{n - 1} \ge m (x^{n + 1} - 1)$.

River Li
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