Let $a,b,c\ge 0: a+b+c=3.$ Find maximum $$P=\frac{1}{\sqrt{a^2-3a+3}}+\frac{1}{\sqrt{b^2-3b+3}}+\frac{1}{\sqrt{c^2-3c+3}}.$$ By $a=b=c=1,$ we got that $P=3$ and we will prove it is desired maximal value.
I tried to use upper bound $$\frac{1}{\sqrt{a^2-3a+3}}\le x.a+y$$
But I'm stuck to find it now. For example $$\frac{1}{\sqrt{a^2-3a+3}}\le \frac{a+1}{2}$$ is saving occuring equality but is wrong in general.
Hope someone can give some advice to help me. Thank you.
For $a=b=c=1$ we obtain a value $3$.
We'll prove that it's a maximal value.
Indeed, by C-S $$\sum_{cyc}\frac{1}{\sqrt{a^2-3a+3}}\leq\sqrt{\sum_{cyc}\frac{1}{(a^2-3a+3)(2a+3)}\sum_{cyc}(2a+3)}$$ and it remains to prove that: $$\sum_{cyc}\frac{1}{(a^2-3a+3)(2a+3)}\leq\frac{3}{5}$$ or $$\sum_{cyc}\frac{1}{((a+b+c)^2-3a(b+c))(3a+b+c)}\leq\frac{27}{5(a+b+c)^3},$$ which is true by BW: https://www.wolframalpha.com/input?i=27-5%28x%2By%2Bz%29%5E3%281%2F%28%28%28x%2By%2Bz%29%5E2-3x%28y%2Bz%29%29%283x%2By%2Bz%29%29%2B1%2F%28%28%28x%2By%2Bz%29%5E2-3y%28x%2Bz%29%29%283y%2Bx%2Bz%29%29%2B1%2F%28%28%28x%2By%2Bz%29%5E2-3z%28y%2Bx%29%29%283z%2By%2Bx%29%29%29%2Cx%3Da%2Cy%3Da%2Bu%2Cz%3Da%2Bv
Suppose $f(x,y,z)=x+y+z-C=0$ And $g(x,y,z)=\mu(x)+\mu(y)+\mu(z)$
By Lagrange Multipliers, we have: $\lambda=d\mu/dx=d\mu/dy=d\mu/dz$
$\mu(x) = \frac{1}{\sqrt{x^2-3x+3}}$
$\frac{d \mu}{dx}=\frac{(2x-3)(-1/2)}{(x^2-3x+3)^{3/2}}$
This implies $x=y=z=1$ and $g(1,1,1)=3$
We also have: $ x^2-3x+3= (x-3/2)^2+3/4$ suggesting a minimum occurs when $x=y=z=3/2$ but that is not allowed by our constraint. Other potential extrema occur on boundaries where $x,y,z$ take on values from $\{0,3\}$ Possibilities are
$z=0. x+y=3\implies y=3-x$
$\frac{1}{\sqrt{(x-3/2)^2+3/4}}+\frac{1}{\sqrt{(3/2-x)^2+3/4}}$
$g(3,0,0)=5/\sqrt{3}<3$
$g(3/2,3/2,0)=5/\sqrt{3}<3$
So max is 3.
Another way.
Let $c=0$.
Thus, since $$a^2-3a+3=\left(a-\frac{3}{2}\right)^2+\frac{3}{4}\geq\frac{3}{4},$$ we obtain: $$\sum_{cyc}\frac{1}{\sqrt{a^2-3a+3}}\leq\frac{4}{\sqrt3}+\frac{1}{3}=\frac{5}{\sqrt3}<3.$$ Let $f(a,b,c,\lambda)=\sum\limits_{cyc}\frac{1}{\sqrt{a^2-3a+3}}+\lambda(a+b+c-3)$ and $abc\neq0$.
Thus, in the inside critical point $(a,b,c)$ should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives $$\frac{2a-3}{\sqrt{(a^2-3a+3)^3}}=\frac{2b-3}{\sqrt{(b^2-3b+3)^3}}=\frac{2c-3}{\sqrt{(c^2-3c+3)^3}}.$$ Now, let $a\neq b$ and $a\neq c$.
Thus, $$\frac{2a-3}{\sqrt{(a^2-3a+3)^3}}=\frac{2b-3}{\sqrt{(b^2-3b+3)^3}}$$ gives $$(2b-3)^2(a^2-3a+3)^3=(2a-3)^2(b^2-3b+3)^3$$ or $$(a-b)(a+b-3)P(a,b)=0,$$ where $P$ is a polynomial of two variables, which gives $P(a,b)=0.$
Also, by the similar way $$\frac{2a-3}{\sqrt{(a^2-3a+3)^3}}=\frac{2c-3}{\sqrt{(c^2-3c+3)^3}}$$ gives $$(a-c)(a+c-3)P(a,c)=0,$$ which gives $P(a,c)=0$.
But, $$P(a,b)-P(a,c)=$$ $$=\tfrac{(2b-3)^2(a^2-3a+3)^3-(2a-3)^2(b^2-3b+3)^3}{(a-b)(a+b-3)}-\tfrac{(2c-3)^2(a^2-3a+3)^3-(2a-3)^2(c^2-3c+3)^3}{(a-c)(a+c-3)}=$$ $$=(b-c)(2a-3)^2(b+c-3)\sum_{cyc}a^2=0,$$ which gives $a=\frac{3}{2}$ or $b=c$.
If $a=\frac{3}{2}$(here we have a saddle point), so $c=\frac{3}{2}-b$ and we need to prove that: $$\frac{1}{\sqrt{b^2-3b+3}}+\frac{1}{\sqrt{b^2+\frac{3}{4}}}+\frac{2}{\sqrt3}\leq3,$$ which is true, but my proof of this statement is very ugly.
But the case $b=c$ says that it's enough to prove our inequality for equality case of two variables.
Let $b=a$ and $c=3-2a$, where $0\leq a\leq\frac{3}{2}.$
Thus, we need to prove that: $$\frac{2}{\sqrt{a^2-3a+3}}+\frac{1}{\sqrt{4a^2-6a+3}}\leq3$$ or $$9(a^2-3a+3)(4a^2-6a+3)\geq\left(\sqrt{a^2-3a+3}+2\sqrt{4a^2-6a+3}\right)^2$$ or $$18a^4-81a^3+140a^2-108a+33\geq2\sqrt{(a^2-3a+3)(4a^2-6a+3)}$$ and since $$18a^4-81a^3+140a^2-108a+33=$$ $$=18\left(a^2-\frac{9}{4}a+1\right)^2+\frac{1}{8}(103a^2-216a+120)>0,$$ it's enough to prove that: $$(18a^4-81a^3+140a^2-108a+33)^2\geq4(a^2-3a+3)(4a^2-6a+3)$$ or $$(a-1)^2(36a^6-252a^5+749a^4-1202a^3+1099a^2-546a+117)\geq0,$$ which is true because $$36a^6-252a^5+749a^4-1202a^3+1099a^2-546a+117=$$ $$=\left(6a^3-21a^2+\frac{41}{2}a-5\right)^2+62\left(a^2-\frac{281}{124}a+\frac{6}{5}\right)^2+$$ $$+\frac{9665a^2-23560a+16864}{6200}>0.$$
Remarks: We may use the Taylor approximation of $f(x) := \frac{1}{\sqrt{x^2-3x+3}}$ around $x = 1$: $\frac{1}{\sqrt{x^2-3x+3}} \sim 1 + \frac12(x - 1) - \frac18(x - 1)^2 + \cdots$. However, we must use it locally. For example, we have $(1 + (x-1)/2)^2 - [f(x)]^2 = \frac{(x^2 + x - 1)(x - 1)^2}{4(x^2 - 3x + 3)}$. Thus, we have $f(x) \le 1 + \frac12(x - 1)$ if $x^2 + x - 1\ge 0$.
WLOG, assume that $a\ge b \ge c$.
Let $$f(x) := \frac{1}{\sqrt{x^2-3x+3}}.$$
Fact 1: $f(x)$ is strictly increasing on $[0, 3/2]$, and strictly decreasing on $[3/2, 3]$.
Fact 2: $f(x) \le 1 + \frac12(x - 1)$ on $[5/8, 3]$.
(Note: $(1 + (x-1)/2)^2 - [f(x)]^2 = \frac{(x^2 + x - 1)(x - 1)^2}{4(x^2 - 3x + 3)}$.)
Fact 3: $f(x) \le 1 + \frac12(x - 1) - \frac18(x - 1)^2$ on $[1, 3]$.
(Note: $(1 + \frac12(x - 1) - \frac18(x - 1)^2)^2
- [f(x)]^2 = \frac{(x^3 - 12x^2 + 30x + 37)(x - 1)^3}{64(x^2 - 3x + 3)}$.)
If $c \le 1/3$, by Fact 1, we have $$P \le f(1/3) + f(3/2) + f(3/2) < 3.$$
If $c \ge 5/8$, by Fact 2, we have $$P \le 1 + \frac12(c - 1) + 1 + \frac12(b - 1) + 1 + \frac12(a-1) = 3.$$
If $1/3 < c < 5/8$ and $b < 1$, by Fact 1, we have $$P \le f(5/8) + f(1) + f(3/2) < 3.$$
If $1/3 < c < 5/8$ and $b \ge 1$, by Fact 3, we have \begin{align*} P &\le \frac{1}{\sqrt{c^2 - 3c + 3}} + 1 + \frac12(b - 1) - \frac18(b - 1)^2 + 1 + \frac12(a - 1) - \frac18(a-1)^2\\ &= \frac{1}{\sqrt{c^2 - 3c + 3}} + 2 + \frac12(1 - c) - \frac{(a - 1)^2 + (b - 1)^2}{8}\\ &\le \frac{1}{\sqrt{c^2 - 3c + 3}} + 2 + \frac12(1 - c) - \frac{(a - 1 + b - 1)^2}{16}\\ &= \frac{1}{\sqrt{c^2 - 3c + 3}} + 2 + \frac12(1 - c) - \frac{(1 - c)^2}{16}\\ &= \frac{1}{\sqrt{c^2 - 3c + 3}} + 3 - \frac{(c + 3)^2}{16}\\ &> 0 \end{align*} where we use $\frac{(c+3)^4}{16^2} > \frac{1}{c^2 - 3c + 3}$ for all $c > 1/3$.
We are done.
Here is my second proof.
Fact 1. If $x, y, z \ge 0$ with
$x^2 + y^2 + z^2 + \frac34 x^2y^2z^2 \le \frac{15}{4}$. Then $x + y + z \le 3$.
(The proof is given at the end.)
Now, let $$x := \frac{1}{\sqrt{a^2-3a+3}}, \quad, y := \frac{1}{\sqrt{b^2-3b+3}}, \quad z := \frac{1}{\sqrt{c^2-3c+3}}.$$
We can prove that $$x^2 + y^2 + z^2 + \frac34 x^2y^2z^2 \le \frac{15}{4}. \tag{1}$$ (The proof of (1) is given at the end.)
By Fact 1, we have $x + y + z \le 3$.
We are done.
$\phantom{2}$
Proof of Fact 1.
We have $x + y \le \sqrt{2(x^2 + y^2)} \le \sqrt{15/2} < 3$. It suffices to prove that $$z^2 \le (3 - x - y)^2. \tag{A1}$$
From $x^2 + y^2 + z^2 + \frac34x^2y^2z^2 \le \frac{15}{4}$, we have $$z^2 \le \frac{15/4 - x^2 - y^2}{1 + 3x^2y^2/4}. \tag{A2}$$
From (A1) and (A2), it suffices to prove that $$\frac{15/4 - x^2 - y^2}{1 + 3x^2y^2/4} \le (3 - x - y)^2. \tag{A3}$$
Let $p = x + y, q = xy$. Then $p < 3$ and $p^2 \ge 4q$.
(A3) is equivalently written as $$3(3 - p)^2q^2 - 8q + 8p^2 - 24p + 21\ge 0$$ or (converting it to a quadratic function in $(p^2/4 - q)$) \begin{align*} &3(3 - p)^2 (q - p^2/4)^2 + \left(8 - \frac{27}{2}p^2 - \frac32p^4 + 9p^3\right)(p^2/4 - q)\\ &\quad + \frac{3}{16}(p^4 - 2p^3 - 3p^2 - 4p + 28)(p-2)^2 \ge 0 \end{align*} which is true since $8 - \frac{27}{2}p^2 - \frac32p^4 + 9p^3 \ge 0$ and $p^4 - 2p^3 - 3p^2 - 4p + 28 \ge 0$ for all $0 \le p \le 3$.
We are done.
$\phantom{2}$
Proof of (1).
We need to prove that \begin{align*} &\frac{1}{a^2-3a+3} + \frac{1}{b^2-3b+3} + \frac{1}{c^2-3c+3} \\[6pt] &\quad + \frac34 \cdot \frac{1}{a^2-3a+3}\cdot \frac{1}{b^2-3b+3}\cdot \frac{1}{c^2-3c+3}\\[6pt] & \le \frac{15}{4}.\tag{B1} \end{align*}
We use the pqr method.
Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$. We have $q \le p^2/3 = 3$.
(B1) is equivalently written as $$15r^2 + (-45q + 123)r + 41q^2 - 222q + 294 \ge 0. \tag{B2}$$
Using degree four Schur inequality, we have $$r \ge \frac{5p^2q - p^4 - 4q^2}{6p} = - \frac29q^2 + \frac52 q - \frac92.\tag{B3}$$
From (B2) and (B3), we have \begin{align*} &15r^2 + (-45q + 123)r + 41q^2 - 222q + 294\\[6pt] \ge{}& 15r \cdot \left(- \frac29q^2 + \frac52 q - \frac92\right) + (-45q + 123)r + 41q^2 - 222q + 294\\[6pt] ={}& \left(- \frac{10}{3}q^2 - \frac{15}{2}q + \frac{111}{2}\right)r + 41q^2 - 222q + r + 294\\[6pt] \ge{}& \left(- \frac{10}{3}q^2 - \frac{15}{2}q + \frac{111}{2}\right)\cdot \left(- \frac29q^2 + \frac52 q - \frac92\right) + 41q^2 - 222q + r + 294\\[6pt] ={}& \frac{1}{108}(80q^2 - 240q + 531)(3 - q)^2 \\[6pt] \ge{}& 0 \end{align*} where we use $- \frac{10}{3}q^2 - \frac{15}{2}q + \frac{111}{2} \ge 0$.
We are done.
Proof.
By your try, we get $$\frac{1}{\sqrt{x^2-3x+3}}\le \frac{x+1}{2}$$ is true for all $x\ge \dfrac{\sqrt{5}-1}{2}.$
WLOG, assuming that $a\ge b\ge c \implies a\ge 1\ge c.$ We divide the original problem into two cases.
Case 1: $c\ge \dfrac{\sqrt{5}-1}{2}$ then $$\sum_{cyc}\frac{1}{\sqrt{a^2-3a+3}}\le \frac{a+b+c+3}{2}=3.$$ Case 2: $c< \dfrac{\sqrt{5}-1}{2}.$ We consider two subcases.
$\bullet b\le 1:$ Notice that $$a^2-3a+3=\left(a-\frac{3}{2}\right)^2+\frac{3}{4}\ge \frac{3}{4}.$$ $$b^2-3b+3=(2-b)(1-b)+1\ge 1.$$ $$c^2-3c+3=(1-c)^2+2-c>\left(1-\frac{\sqrt{5}-1}{2}\right)^2+2-\frac{\sqrt{5}-1}{2}=6-2\sqrt{5}.$$ It implies $$\sum_{cyc}\frac{1}{\sqrt{a^2-3a+3}}< \frac{2}{\sqrt{3}}+1+\frac{1}{\sqrt{5}-1}<3.$$ $\bullet b\ge 1 \implies 2\ge a\ge b\ge 1.$ Consider the function $f(x)=\dfrac{1}{\sqrt{x^2-3x+3}}$ for $x\in [1;2]. $
Notice that $$f''(x)=\frac{8x^2-24x+15}{4(x^2-3x+3)^{\frac{3}{2}}}<0, \forall x\in[1;2]$$ Thus, by Jensen inequality $$f(a,)+f(b)\le 2f\left(\frac{a+b}{2}\right)=2f(t)=\frac{2}{\sqrt{t^2-3t+3}}.$$ The rest is proving $$\frac{2}{\sqrt{t^2-3t+3}}+\frac{1}{\sqrt{(3-2t)^2-3(3-2t)+3}}\le 3,$$or $$\frac{2}{\sqrt{t^2-3t+3}}+\frac{1}{\sqrt{4t^2-6t+3}}\le 3,$$which is true for all $1\le t\le 2.$
The proof is done. The maximal value is $3$ achieve at $a=b=c=1.$
Here is my third proof.
We use the so-called isolated fudging.
After homogenization, it suffices to prove that, for all $a, b, c \ge 0$ with $a + b + c > 0$, $$\sum_{\mathrm{cyc}} \frac{a + b + c}{\sqrt{9a^2 - 9a(a + b + c) + 3(a + b + c)^2}} \le 3. \tag{1}$$
It suffices to prove that, for all $a, b, c \ge 0$ with $a + b + c > 0$, $$\frac{a + b + c}{\sqrt{9a^2 - 9a(a + b + c) + 3(a + b + c)^2}} \le \frac{10a^2 + 7b^2 +7c^2 + 2a(b + c) - 4bc}{8(a^2 + b^2 + c^2)}. \tag{2}$$
We split into two cases.
Case 1. $a = 0$
(2) becomes $$\frac{1}{\sqrt{3}} \le \frac{7b^2 +7c^2 - 4bc}{8(b^2 + c^2)}$$ which is true (easy).
Case 2. $a > 0$
WLOG, assume that $a = 1$. (2) is written as $$\frac{1 + b + c}{\sqrt{9 - 9(1 + b + c) + 3(1 + b + c)^2}} \le \frac{10 + 7b^2 +7c^2 + 2(b + c) - 4bc}{8(1 + b^2 + c^2)}. \tag{3}$$
Let $p = b + c, q = bc$. Then $p^2 \ge 4q$.
(3) is written as $$\frac{1 + p}{\sqrt{3p^2 - 3p + 3}} \le \frac98 - \frac{2p^2 - 2p - 1}{8(p^2 - 2q) + 8}. \tag{4}$$
If $2p^2 - 2p - 1 \ge 0$, using $p^2 \ge 4q$, it suffices to prove that $$\frac{1 + p}{\sqrt{3p^2 - 3p + 3}} \le \frac98 - \frac{2p^2 - 2p - 1}{8(p^2 - 2\cdot \frac{p^2}{4}) + 8}$$ which is true for all $p \ge 0$ (not difficult).
If $2p^2 - 2p - 1 < 0$, it suffices to prove that $$\frac{1 + p}{\sqrt{3p^2 - 3p + 3}} \le \frac98 - \frac{2p^2 - 2p - 1}{8(p^2 - 2\cdot 0) + 8}$$ which is true for all $p \ge 0$ (not difficult).
We are done.