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Problem. Find the minimal $k$ value such that $$\frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{\sqrt{a+b+c+k\cdot abc}}\le 1+\sqrt{2}$$ holds for all $a,b,c\ge 0: ab+bc+ca=1.$

I'm not sure my prediction is correct 100%. Hope you can point out my mistakes.

By plugging $a=b=c=\dfrac{\sqrt{3}}{3},$ we get $k\ge 153-108\sqrt{2}:= k_0.$

Now, we'll prove $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le (1+\sqrt{2}) \sqrt{a+b+c+k_0\cdot abc}.$$

It means that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le (1+\sqrt{2}) \sqrt{a+b+c+k_0\cdot \frac{abc}{ab+bc+ca}}.$$ I thought of squaring both side of the last inequality but nothing interesting left.

I hope that my $k_0$ is right answer. In case it's true, you can share your idea and comment how to full th remain part.

Thanks for interest.

River Li
  • 49,125

1 Answers1

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Some thoughts.

Remark: It is similar to my answer here.

It suffices to prove that $$(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2 \le (3 + 2\sqrt 2)(a + b + c + k_0abc). \tag{1}$$

By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\\[6pt] \le{}& \left(\sum_{\mathrm{cyc}} \frac{a + b}{\frac12a + \frac12b+ \frac{3 - 2\sqrt{2}}{2}c(a + b) - (3 - 2\sqrt{2})ab + \frac13k_0 abc}\right) \\[6pt] &\qquad \times \left(\sum_{\mathrm{cyc}} \left(\frac12a + \frac12b+ \frac{3 - 2\sqrt{2}}{2}c(a + b) - (3 - 2\sqrt{2})ab + \frac13k_0 abc\right)\right)\\[6pt] ={}& \left(\sum_{\mathrm{cyc}} \frac{a + b}{\frac12a + \frac12b+ \frac{3 - 2\sqrt{2}}{2}c(a + b) - (3 - 2\sqrt{2})ab + \frac13k_0 abc}\right) (a + b + c + k_0abc). \end{align*}

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{a + b}{\frac12a + \frac12b+ \frac{3 - 2\sqrt{2}}{2}c(a + b) - (3 - 2\sqrt{2})ab + \frac13k_0 abc} \le 3 + 2\sqrt{2}. \tag{2}$$ (2) is true which is verified by Mathematica - a Computer Algebra System (CAS). I think we can use the pqr method.

River Li
  • 49,125