# Find $k=constant$ such that $f(a,\,b,\,c\,)=\frac{3a+2b}{\sqrt{5a^2-ab+b^2}}+\frac{3b+2c}{\sqrt{5b^2-bc+c^2}}\leqq f(a,k\,a+\sqrt[3]{abc}-k\,c,c\,)$

Give $$3$$ positve numbers $$a,\,b,\,c$$ such that $$abc= 1$$ , prove: $$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\leqq f\left ( a,\,1,\,c \right )$$ Because the inequality is homogeneous, so we can write: $$f\left ( a,\,b,\,c \right )= \frac{3\,a+ 2\,b}{\sqrt{5\,a^{\,2}- ab+ b^{\,2}}}+ \frac{3\,b+ 2\,c}{\sqrt{5\,b^{\,2}- bc+ c^{\,2}}}\leqq f\left ( a,\,\sqrt[3\,]{abc},\,c \right )$$ I tried to break the square root by using: $$\sqrt{A}- B= \frac{A- B^{\,2}}{\sqrt{A}+ B}= \frac{A- B^{\,2}}{\frac{A- C^{\,2}}{\sqrt{A}+ C}+ B+ C}= ...$$ This is hard, I solved easier problem, we have: $$\sqrt{5\,a^{\,2}- ab+ b^{\,2}}- \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}}= \frac{\left ( 23\,a- 8\,b \right )^{\,2}}{140\left [ \sqrt{5\,a^{\,2}- ab+ b^{\,2}}+ \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}} \right ]}$$ $$\sqrt{5\,b^{\,2}- bc+ c^{\,2}}- \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}}= \frac{\left ( 23\,b- 8\,c \right )^{\,2}}{140\left [ \sqrt{5\,b^{\,2}- bc+ c^{\,2}}+ \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}} \right ]}$$ Let $$g\left ( a, b \right )= \sqrt{5\,a^{\,2}- ab+ b^{\,2}}+ \left ( 3\,a+ 2\,b \right )\sqrt{\frac{19}{40}}$$ . I tried to prove: $$\left ( 8\,c- 23\,b \right )\left \{ g\left ( \frac{b^{\,2}}{c},\,b \right )- \left [ \sqrt{5\,b^{\,2}- bc+ c^{\,2}}+ \left ( 3\,b+ 2\,c \right )\sqrt{\frac{19}{40}} \right ] \right \}\geqq 0$$ But without success, I think: $$0\leqq f\left ( a,\,\sqrt[3\,]{abc},\,c \right )- f\left ( a,\,b,\,c \right )= \left ( \sqrt[3\,]{abc}- b \right )A\Leftrightarrow b^{\,2}\geqq ac\Leftrightarrow a\leqq \frac{b^{2}}{c}\Leftrightarrow g\left ( a, b \right )\leqq g\left ( \frac{b^{2}}{c}, b \right )$$ Edit: I find $$k= constant$$ such that: $$f(\,a,\,b,\,c\,)\leqq f(\,a,\,k\,a+ \sqrt[3\,]{abc}- k\,c,\,c\,)$$. Thank you a real lot!

The inequality in your title is incorrect. There is no $$k$$ for which this is true.

Let us set $$a = c$$ in your inequality. Then we have

$$f(a,b,a) \leq f(a, (a^2b)^{1/3}, a).$$

This implies

$$f(a,b,a) \leq f(a, (a^2b)^{1/3}, a) \leq f(a, (a^8 b)^{1/9}, a) \leq f(a, (a^{26} b)^{1/27}, a) \leq ... \leq f(a, a, a).$$

This implies that, when $$a$$ is fixed, $$f(a,b,a)$$ is maximized at $$b = a$$.

However, plugging in $$a=1$$, it seems that while the point $$b=a$$ is indeed an inflection point of $$f(a,b,a)$$, the point $$b=a$$ is a local minimum and not a local maximum, a contradiction. See the plot below.

NOTE: the original question was whether the inequality was correct for $$k=1$$. Here is my answer to that.

Your inequality is incorrect: $$f(1,3,1/3) = 4.169$$ and $$f(1,1,1/3) = 3.914.$$

However, your inequality does seem to hold for $$a \leq 1/2$$, so you should look and see for what $$(a,b,c)$$ you actually need it to work.

ADDED NOTE: it's actually not true for $$a \leq 1/2$$, as plotting the function $$f(a, b, a)$$ for $$a=\frac{1}{2}$$ shows.