Inequality. Right?

Put P under a common denominator:

$P = \frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca}+ \frac{c^{2}}{2c^{2}+ab} \\ = \frac{a^{2}(2b^{2}+ca)(2c^{2}+ab) + b^{2}(2a^{2}+bc)(2c^{2}+ab) + c^{2}(2a^{2}+bc)(2b^{2}+ca)}{(2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab)} \\ = \frac{a^{2}((2bc)^{2} + 2a(b^{3} + c^{3}) + a^{2}bc) + b^{2}((2ac)^{2} + 2b(a^{3} + c^{3}) + ab^{2}c) + c^{2}((2ab)^{2} + 2c(a^{3} + b^{3}) + abc^{2})}{((2ab)^{2} + 2c(a^{3} + b^{3}) + abc^{2}))(2c^{2} + ab)} \\ = \frac{12(abc)^{2} + 2a^{3}(b^{3} + c^{3}) + a^{4}bc + 2b^{3}(a^{3} + c^{3}) + ab^{4}c + 2c^{3}(a^{3} + b^{3}) + abc^{4}}{8(abc)^{2} + 4((ab)^{3} + (bc)^{3} + (ac)^{3}) + 2abc(a^{3}+b^{3}+c^{3})} \\ = \frac{12(abc)^{2} + 4((ab)^{3} + (ac)^{3} + (bc)^{3}) + abc(a^{3} + b^{3} + c^{3})}{9(abc)^{2} + 4((ab)^{3} + (bc)^{3} + (ac)^{3}) + 2abc(a^{3}+b^{3}+c^{3})} \\$ [1]

It is given that for real numbers a, b, c $a+b+c=0 \\ (a+b+c)^{2} = 0^{2} \Rightarrow a^{2}+ b^{2}+c^{2} = -2(ab+ac+bc) \\ (a+b+c)^{3} = 0^{3} \\ a^{3}+ b^{3}+c^{3} + 3abc(a+b+c) + 3(a^{2}b + a^{2}c + ab^{2} + b^{2}c + ac^{2} + bc^{2}) + 6abc = 0 \\ a^{3}+ b^{3}+c^{3} = -[3abc(0) + 3(a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b) + 6abc] \\ a^{3}+ b^{3}+c^{3} = -3(a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b) - 6abc \\$ [2]

Since $(a^{2}+b^{2}+c^{2})(a+b+c)$ are defined, $(a^{2}+b^{2}+c^{2})(a+b+c) = a^{3}+b^{3}+c^{3} + a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b \\ -2(ab+ac+bc)(0) = 0 = a^{3}+b^{3}+c^{3} + a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b \\ a^{2}b + a^{2}c + b^{2}a + b^{2}c + c^{2}a + c^{2}b = -(a^{3}+b^{3}+c^{3})$ [3]

Combining [1] and [2] gives $a^{3}+ b^{3}+c^{3} = 3(a^{3} + b^{3} + c^{3}) - 6abc => a^{3} + b^{3} + c^{3} = 3abc$

When this is put back into equation [1], we get $P = \frac{12(abc)^{2} + 4((ab)^{3} + (ac)^{3} + (bc)^{3}) + abc(3abc)}{9(abc)^{2} + 4((ab)^{3} + (bc)^{3} + (ac)^{3}) + 2abc(3abc)} = \frac{15(abc)^{2} + 4((ab)^{3} + (ac)^{3} + (bc)^{3})}{15(abc)^{2} + 4((ab)^{3} + (bc)^{3} + (ac)^{3})} = \boxed{1}$

$P =\frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca}+ \frac{c^{2}}{2c^{2}+ab} \\ =\dfrac {\sum_{cyc}a^2*(2b^2+ca)(2c^2+ab)} {(2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab)} \\ \color{#3D99F6}{(2b^2+ca)(2c^2+ab)=4*b^2*c^2+2a(b^3+c^3) + a*abc.}\\ \ \ \ \ \text{This will simplify our calculations.} \\ 1st \ Cycle\ \ \ a^2*(2b^2+ca)(2c^2+ab)=4(a^2*b^2*c^2) + 2a^3(b^3+c^3)+a^3*abc\\ 2nd \ Cycle\ \ b^2*(2c^2+ab)(2a^2+bc)=4(a^2*b^2*c^2) + 2b^3(c^3+a^3)+b^3*abc\\ 3rd \ Cycle\ \ \ c^2*(2a^2+bc)(2b^2+ca)=4(a^2*b^2*c^2) + 2c^3(a^3+b^3)+c^3*abc\\ \therefore\ \sum_{cyc}a^2*(2b^2+ca)(2c^2+ab)\\ =12(a^2*b^2*c^2)+4(a^3*b^3+b^3*c^3+c^3*a^3) +( a^3+b^3+c^3)*abc,\\ But\ a^3+b^3+c^3= 3abc.\\ \therefore\ \ \sum_{cyc}= \color{#3D99F6}{15(a^2*b^2*c^2)+4(a^3*b^3+b^3*c^3+c^3*a^3) } \\ (2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab)={\Large 2*}(4(a^2*b^2*c^2) + 2a^3(b^3+c^3)+a^3*abc)\\ \text{We have seen from 1st cycle that }\\ a^2*(2b^2+ca)(2c^2+ab)=4(a^2*b^2*c^2) + 2a^3(b^3+c^3)+a^3*abc.\\ bc*\color{#3D99F6}{(2b^2+ca)(2c^2+ab)=bc*\{4*b^2*c^2+2a(b^3+c^3) + a*abc.\}}\\ =4b^3*c^3+2abc(b^3+c^3)+a^2*b^2*c^2.\\ \therefore\ \ (2a^{2}+bc)(2b^{2}+ca)(2c^{2}+ab)\\ =\{8(a^2*b^2*c^2) + 4a^3(b^3+c^3)+2a^3*abc\} + \{4b^3*c^3+2abc(b^3+c^3)+a^2*b^2*c^2\}. \\ ={ \Large (}8(a^2*b^2*c^2)+\{2a^3*abc+ 2abc(b^3+c^3)\}+a^2*b^2*c^2 { \Large )} \ \ + \ \ { \Large (} 4a^3(b^3+c^3)+4b^3*c^3{ \Large )} \\ = \color{#3D99F6}{15(a^2*b^2*c^2)+4(a^3*b^3+b^3*c^3+c^3*a^3) } \\ \therefore\ \ P= \color{#D61F06}{1} \\\ \ \ \\ \\$
Since the calculations are little involved I have given them in details for those who need.
Note the following that makes calculations easy. 1. Before first cycle, I have calculated multiplication only of two groups. 2. Once we get the first cycle, the remaining simply rotate the variables, that is, for a, put b, for b, put c , and for c, put a. Please take care to do it at the spot every time. If you try to change all say a's at a time, you will land into a mess. 3. see I have not calculated denominator fully. Previous calculations have been used.