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# Understand the problem

Let a, b, c be positive real numbers such that a + b + c = 1. Prove that $$\frac {a} {a^2 + b^3 + c^3} + \frac {b}{ b^2 + c^3 + a^3 } + \frac {c} { c^2 + a^3 + b^3 } \leq \frac{1}{5abc}$$

##### Source of the problem

Regional Math Olympiad, 2019 Problem 3

Inequality
6/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

The clue: Number 5 in the right-hand side! We will be applying AM-GM inequality. But first, to get the 5 on the right, we need 5 terms in the left (or bunch of five terms).

Try to use a+b+c =1 to cook it up!

In the first term’s denominator, we have $a^2 + b^3 + c^3$. Multiply 1 to $a^2$ that is multiply by a + b + c (nothing changes because, multiplying by does not change anything). Hence we have $a^2 \cdot 1 + b^3 + c^3 = a^2 ( a + b + c) + b^3 + c^3$ Expanding we have  $a^3 + b^3 + c^3 + a^2 b + a^2 c$ Now apply AM – GM inequality to this we have $$\frac{ a^3 + b^3 + c^3 + a^2 b + a^2 c}{5} \geq (a^3 \cdot b^3 \cdot c^3 \cdot a^2 b \cdot a^2 c )^{1/5}$$ Therefore we have  $$a^3 + b^3 + c^3 + a^2 b + a^2 c \geq 5 \cdot (a^7 b^4 c^4)^{1/5}$$ Taking the reciprocal we have and noting that the left hand side is still $a^2 + b^3 + c^3$ we have  $$\frac {1} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{7/5} b^{4/5} c^{4/5} }$$ Multiplying the numerator and denominator by a we have the desired expression in the left. $$\frac {a} {a^2 + b^3 + c^3} \leq \frac {a}{5 \cdot (a^{7/5} b^{4/5} c^{4/5} }$$ Simplifying $$\frac {a} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} }$$ Now try computing the same for the other two terms on the left.
$$\frac {a} {a^2 + b^3 + c^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} }$$ $$\frac {b} {b^2 + c^3 + a^3} \leq \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} }$$ $$\frac {c} {c^2 + a^3 + b^3} \leq \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }$$ Adding we have  $$\frac {a} {a^2 + b^3 + c^3} + \frac {b} {b^2 + c^3 + a^3} + \frac {c} {c^2 + a^3 + b^3} \leq \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }$$ Now apply AM- GM Inequality one more time to the left hand term.
$$\frac{ \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }}{3} \leq ( \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } \cdot \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } \cdot \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} })^{1/3}$$ Simplifying  we have  $$\frac{ \frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} }}{3} \leq ( \frac {1}{5^3 \cdot (a^{10/5} b^{10/5} c^{10/5} } )^{1/3}$$ Simplifying further we have $$\frac {1}{5 \cdot (a^{2/5} b^{4/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{2/5} c^{4/5} } + \frac {1}{5 \cdot (a^{4/5} b^{4/5} c^{2/5} } \leq 3 \cdot \frac {(abc)^{1/3}}{5abc}$$ Finally we know that $3 \cdot (abc)^{1/3} \leq 1$. Why? Apply AM-GM to a, b, c $$\frac{a+b+c}{3} \geq (abc)^{1/3}$$ Since a + b + c = 1 we have the result.

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