Hint

Yeah.

$$ 2a^2+2b^2+2c^2+2a-1\geq a^2+2a+4b+4c+2a-6=a^2+6$$$$ \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1}\leq 1,... $$

Nice

Functional_equation wrote: $a,b,c$ are positive numbers.$a+b+c=3$ Prove that: $\sum \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1}\leq 3 $ Applying Cauchy-Schwarz inequality, we have $$2b^2+2c^2 \enspace = \enspace (1+1) (b^2+c^2) \enspace \geq (b+c)^2$$ and using the condition $a+b+c =3$, this is equivalent to $$ \iff \quad 2b^2+2c^2 \enspace \geq \enspace (3-a)^2 \enspace = \enspace a^2-6a+9$$ Adding $(2a^2+2a-1)$ to both sides, this is equivalent to $$ \iff \quad 2a^2+2b^2+2c^2+2a-1 \enspace \geq \enspace 3a^2 -8a +8$$ $$ \iff \quad \frac{a^2+6}{3a^2-8a+8} \enspace \geq \enspace \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1} \quad \quad \quad \quad \quad (1)$$ Now observe that the inequality $$1 \enspace \geq \enspace \frac{a^2+6}{3a^2-8a+8} \quad \quad \quad \quad \quad (2)$$ is equivalent to $$ \iff \quad 3a^2-8a+8 \enspace \geq \enspace a^2+6$$ $$ \iff \quad 2(a-1)^2 \enspace \geq \enspace 0 $$ which is obviously true. Combining $(1)$ and $(2)$, we obtain $$1 \enspace \geq \enspace \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1} \quad \quad \quad \quad \quad \quad \quad (3) $$ Similarly we obtain $$1 \enspace \geq \enspace \frac{b^2+6}{2a^2+2b^2+2c^2+2b-1} \quad \quad \quad \quad \quad \quad \quad (4) $$ and $$1 \enspace \geq \enspace \frac{c^2+6}{2a^2+2b^2+2c^2+2c-1} \quad \quad \quad \quad \quad \quad \quad (5) $$ Adding the inequalities $(3)$, $(4)$ and $(5)$ we get the desired inequality $$3 \enspace\geq \enspace\frac{a^2+6}{2a^2+2b^2+2c^2+2a-1} \enspace +\enspace \frac{b^2+6}{2a^2+2b^2+2c^2+2b-1} \enspace+\enspace\frac{c^2+6}{2a^2+2b^2+2c^2+2c-1}$$

Functional_equation wrote: $a,b,c$ are positive numbers.$a+b+c=3$ Prove that: $\sum \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1}\leq 3 $ Similiar $a,b,c$ are positive numbers.$a+b+c=3$ Prove that: $\sum \frac{122-17a^2}{2a^2+2b^2+2c^2+2a-1}\leq 45 $

lemma: a*2+b*2+c*2 >=ab+ac+bc so a*2+b*2+c*2>=3

$\spadesuit \ \color{blue}{\textbf{Solution:}}$ By homogenizing the inequality, it's equivalent to proving that \[\sum_{cyc} \frac{a^2+\frac{2}{3}(a+b+c)^2}{2(a^2+b^2+c^2)+\frac{2a(a+b+c)}{3}-\frac{(a+b+c)^2}{3}}\le 3.\]we now show that \[\frac{a^2+\frac{2}{3}(a+b+c)^2}{2(a^2+b^2+c^2)+\frac{2a(a+b+c)}{3}-\frac{(a+b+c)^2}{3}} \le 1.\]Simplifying the expressions, we will end up proving \[8a^2+11b^2+11c^2\ge 8ab+14bc+8ac\]which follows from AM-GM, \[4a^2+4b^2\ge 8ab\]\[4a^2+4c^2\ge 8ac\]\[7b^2+7c^2 \ge 14bc.\]The other 2 cases are similar, so \[\sum_{cyc} \frac{a^2+\frac{2}{3}(a+b+c)^2}{2(a^2+b^2+c^2)+\frac{2a(a+b+c)}{3}-\frac{(a+b+c)^2}{3}}\le \sum_{cyc} 1 = 3. \quad \blacksquare\]

Lemma : $3(a^2+b^2+c^2) \geq (a+b+c)^2 \to a^2+b^2+c^2 \geq 3 $ Then we can write : $\sum \frac{a^2+6}{2a^2+2b^2+2c^2+2a-1}\leq \sum \frac{a^2+6}{2a+5} \leq 3 $ It is same to : $P(x) : a^2-6a-9 \leq 0 $ . And it is true because $\Delta \geq 0$

$a^2 + 6 \leq ( 2a^2 + 2b^2 + 2c^2 + 2a - 1) $