I just learned today that there can be a tighter result than AM-GM (Arithmetic Mean - Geometric Mean) inequality. In particular:

Let $a, b > 0$ then \begin{equation} \label{1}\tag{1} \dfrac{a+b}{2} - \sqrt{ab} \geq \dfrac{1}{16 \max \left\lbrace a , b \right\rbrace} \left( a - b \right) ^{2} . \end{equation}

When the RHS of \eqref{1} is $0$ then we get the classical AM-GM inequality. This result is obviously better since this RHS could be greater than $0$ in general.

I wonder if we can get some similar result for Cauchy - Schwarz inequality. For example, just a particular case, is there some $\mathcal{E} \geq 0$ such that \begin{equation} \left( a + b \right) ^{2} \leq 2 \left( a^{2} + b^{2} \right) - \mathcal{E} . \end{equation}


I think you mean $\mathcal{E}\geq0$, otherwise $a=b$ will give a counterexample.

For example: $$2(a^2+b^2)-\frac{(a-b)^4}{a^2+b^2}\geq(a+b)^2.$$

For three variables we have the following stronger than C-S inequality.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$3(a^2+b^2+c^2)-(a+b+c)^2\geq\frac{25(a-b)^2(a-c)^2(b-c)^2}{a^4+b^4+c^4}.$$

If we'll change $25$ on $26$ we'll get a wrong inequality.

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    $\begingroup$ Isn't the question about Cauchy-Schwartz ? $\endgroup$
    – user65203
    Oct 5 '17 at 13:04
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    $\begingroup$ @Yves Daoust Thank you! I'll fix. $\endgroup$ Oct 5 '17 at 13:12
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    $\begingroup$ To prove it, one method you could use is notice that $$2(a^2+b^2)-\frac{(a-b)^4}{a^2+b^2}-$$ $$-(a+b)^2=\frac{2ab(a-b)^2}{a^2+b^2}\ge 0$$ for all $a,b$, where $a,b$ have the same non-zero sign or exactly one of them is $0$. $\endgroup$
    – user236182
    Oct 5 '17 at 14:26
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    $\begingroup$ @user236182 Yes of course! My second inequality is much more harder. $\endgroup$ Oct 5 '17 at 14:34

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