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

Question

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!

Answer 1

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.