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$\Omega$ is a compact subset, $n\ge 2$ is a integer, $a,b \in C^\infty(\Omega)$, C is a positive constant. If $$ \frac{C^2}{2(n-1)}+(a+b)C\le \frac{a^2}{n-1} $$ How to show $$ C\le\max\{4(n-1)(||-b||_\infty+||a||_\infty), \sqrt 8 ||a||_\infty\} $$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Picture below is the origin of this question, I can't get the last inequation. enter image description here

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    When you write $b$ in the desired inequality do you really mean $\Vert b \Vert_\infty$? Otherwise it's a bit strange that $b$ appears as a function but $a$ appears only in norms. – 2017-01-01
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    @Glitch I think you are right , I have add the origin of my question and review it . Found there is something wrong,from the pictures, the (1.12) is right only when $x=x_0$ , so there will be not result like the last inequation. So , I think $-b$ should be altered $||-b||_\infty$. – 2017-01-02
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    @Glitch I am very sorry. I read the back matter, and the theorem 1 must be same with picture above, so , $-b$ can't replaced by $||-b||_\infty$. I think there are something I don't understant in proof of theorem 1. – 2017-01-02

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First, for any $a\le b+c$, we have $a\le \max\{2b, 2c\}$. So, for 1.12, we have $$ f^2(x_0)\le 2(n-1) \max\{2(-F_u-(n-1)K-\frac{F(u)}{\mu\sup u -u})f(x_0)~,~ \frac{2}{n-1}(\frac{F(u)}{\mu\sup u -u})^2\} $$ Then, the last equation is easy to get, and $\sqrt 8 $ can be replaced by $2$.