If a,b,c are positive real numbers and P = (b^2+c^2)/(b+c) + (c^2+a^2)/(c+a) + (a^2+b^2)/(a+b) , ...?

then only one of the following statements is always true. Which one is it?


A) 0 %26lt;= P %26lt; a;


B) a %26lt;= P %26lt; a+b;


C) a+b %26lt;= P %26lt; a+b+c;


D) a+b+c %26lt;= P %26lt; 2(a+b+c);


Explain your answer...

If a,b,c are positive real numbers and P = (b^2+c^2)/(b+c) + (c^2+a^2)/(c+a) + (a^2+b^2)/(a+b) , ...?
P = (b²+c²)/(b+c) + (c²+a²)/(c+a) + (a²+b²)/(a+b)


= b+c - 2bc/(b+c) + c+a - 2ac/(c+a) + a+b - 2ab/(a+b)


= 2(a+b+c) - (2bc/(b+c) + 2ac/(a + c) + 2ab/(a + b))





2bc/(b + c) is the harmonic mean of b and c


The harmonic mean ≤ the arithmetic mean


The harmonic mean %26gt; 0 (a, b and c are all positive numbers)





2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ (b+c)/2 + (a+c)/2 + (a+b)/2


(The sum of three harmonic means must be less than or equal to the sum of the corresponding arithmetic means)








2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ 2(a+b+c)/2


2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) ≤ (a+b+c)





2bc/(b+c) + 2ac/(a+c) + 2ab/(a+b) %26gt; 0


(Each harmonic mean is greater than 0, so the sum must be greater than 0)





P %26lt; 2(a+b+c)





P ≥ 2(a+b+c) - (a+b+c)


P ≥ (a+b+c)





(a+b+c) ≤ P %26lt; 2(a+b+c)





D