My own problems – all in one place

\boxed{1}
Prove that for all a, b, c\in\mathbb{R}^{ + }, such that a + b + c = 1, we have:
\frac {a(1 - a)^2}{bc} + \frac {b(1 - b)^2}{ca} + \frac {c(1 - c)^2}{ab}\ge 4
(Posed Here)

\boxed{2}
Prove that for x, y, z\in \mathbb{R}^{ + } such that x + y + z = 1,
\frac {1 - x}{2yz} + \frac {1 - y}{2zx} + \frac {1 - z}{2xy}\ge

\ge \frac {1}{\sqrt {x^2 + y^2 - xy}} + \frac {1}{\sqrt {y^2 + z^2 - yz}} + \frac {1}{\sqrt {z^2 + x^2 - zx}} >

> \frac {1}{1 - x} + \frac {1}{1 - y} + \frac {1}{1 - z}\ge \frac {9}{2}
(Posted Here)

\boxed{3}
Prove that , for all a, b, c\in\mathbb{R}^{ + } such that a + b + c = 3, we have:
\sqrt {\left( \sum_{cyc}\frac {a^2 + 2c^2}{b + c} \right) \left(\sum_{cyc} \frac {a}{(b + c)^2}\right)}\ge\frac {3\sqrt {3}}{...
(Posted Here)[/hide]

\boxed{4}
For a, b, c\in \mathbb{R} such that a + b + c\ge1, prove that:
4(a + b + c) + \frac {a^2}{b + c} + \frac {b^2}{c + a} + \frac {c^2}{a + b}\ge\frac {9}{2}
(Posted Here)

\boxed{5}
Prove that, for a,b,c\ge 0 satisfying \sum_{cyc}\frac {1}{1 + a^2} = \frac 12 we always have:
\sum_{cyc}\frac1{a^3 + 2}\leq \frac {1}{3}
(Posted Here; See post # 75)

\boxed{6}
For positive reals a,b,c we have:
\frac {(a + b)(b + c)(c + a)}{8abc}\ge\frac {\sqrt {\prod_{sym}(a^{2} + b)}}{(a + bc)(b + ca)(c + ab)}
(Posted here)

\boxed{7}
For positive reals x_i satisfying \sum_{i = 1}^n x_i = n Find P_{min} given k\in\mathbb N:
P = \frac {x_{1}^{k}}{x_{n} + x_{1}} + \frac {x_{2}^{k}}{x_{1} + x_{2}} + \cdots + \frac {x_{n}^{k}}{x_{1} + x_{n - 1}}
(Posted here)

\boxed{8}
For positive reals a,b,c prove that :
\sum\frac {a^{5}}{b^{2}c^{2} + ab^{2}c}\geq\frac {a + b + c}{2}
(I jumbled up other three inequalities of mine with this one; Posted here)

\boxed{9}
For Positive reals x,y,z prove that:
\sum\frac {x^{5}}{x^{3}yz + yz^{4}}\geq\frac {3}{2}
(Posted Here)

\boxed{10}
For positive reals a,b,c satisfying a + b + c = 3 prove that:
12 - \sum_{cyc}\left[\frac {a}{bc}(13b^{2} - 3a^{2})\right]\geq 6\sum_{cyc}\left[\frac {b^{2} + c}{c}\right]
(Posted Here)

\boxed{11}
Prove that , for a,b,c > 0 satisfying a^{4} + b(b^{2} + 1) + 4c = 7:
\frac {(a + b)^{2}}{b(b + c)^{2}} + \frac {(b + c)^{2}}{c(c + a)^{2}} + \frac {(c + a)^{2}}{a(a + b)^{2}}\geq a + b + c
(Posted here)

\boxed{12}
For a,b,c > 0; a + b + c = 1 Prove that:
\sum_{cyc}\frac {a^3}{(1 - a)^2}\geq \frac 14
(Posted here; Post #138)

\boxed{13}
Prove that for all positive reals a,b,c we always have:
\frac {a^2 + bc}{a + b} + \frac {b^2 + ca}{b + c} + \frac {c^2 + ab}{c + a}\geq 3\sqrt [3]{abc}
(Posted Here)

\boxed{14}
For a,b,c > 0, show that
\frac a{5a + 4b + 3c} + \frac b{5b + 4c + 3a} + \frac c{5c + 4a + 3b}\le \frac 19 \sum \frac a{a + b} + \frac {10}{27}
(Posted here)

\boxed{15}
Given a,b,c are the sides of a triangle, show that we have
\sum_{cyc}\sqrt {\frac {a}{c + a - b}}\ge 3
(Not posted anywhere yet)

\boxed{16}
For all x,y,z > 0 we have that
\frac {x + y + z}{\sqrt {3}}\geq \frac {xy + yz + zx + 2\sqrt {xyz}\left(\sqrt {x} + \sqrt {y} + \sqrt {z}\right)}{\sqrt {x^2...
(Not posted anywhere yet;)
Note

\boxed{17}
Given a,b,c > 0 such that ab + bc + ca = abc, prove that we have
\frac 1{a + 3b + 2c} + \frac1{b + 3c + 2a} + \frac 1{c + 3a + 2b}\leq \frac 16
(Posted Here)

\boxed{18}
For all a,b,c > 0, show that
\sqrt {\frac {a^2 + bc}{b(c + a)}} + \sqrt {\frac {b^2 + ca}{c(b + a)}} + \sqrt {\frac {c^2 + ab}{a(b + c)}}\geq 3
(Not posted Anywhere Yet)

\boxed{19}
(Sayan Mukherjee) For positive reals a,b,c satisfying a + b + c = 3; Prove that we have:
\displaystyle \sum_{cyc}\frac {a}{bc}\left(a^2 + 7b^2\right) \ge 4\sum_{cyc}\frac {ab}{c} + 2\sum_{cyc}\frac {b^2 + c}{c}.
(Not posted Anywhere Yet)

\boxed{20}
Given positive reals a,b,c; Prove that we have
\displaystyle\frac {(a + b + c)^2}{ab + bc + ca}\geq \frac 49 \cdot \frac {(a + b + c)^2}{ab + bc + ca}\cdot \left(\frac b{a ...
(Not posted Anywhere Yet)
(Proposed in the contest named War_inequality_Secondary at maths.vn as a part of Team 1)

\boxed{21}
For all a,b,c\geq 0, show that we have
\displaystyle 128(a + b + c)(a - b)(b - c)(c - a)\leq 9(a^2 + b^2 + c^2)^2
Note
(Not posted Anywhere Yet)
\boxed{22}
Given a,b,c > 0, show that we have
\frac {a^6}{(a + b)(a^4 + a^2b^2 + b^4)} + \frac {b^6}{(b + c)(b^4 + b^2c^2 + c^4)} + \frac {c^6}{(c + a)(c^4 + c^2a^2 + a^4)...
(Posted here)
\boxed{23}
For positive reals a,b,c show that we have
\sum_{cyc}\frac 1{(2a^2 + bc)(b + c)^2}\leq \frac 9{4(ab + bc + ca)^2}.
(Made while proving problem wrongly here)
Remark.
This is actually weaker than a problem by Tran Nguyen Quoc Cuong (mathstarofvn), which is on proving that
\sum_{cyc}\frac 1{(2a^2 + bc)}\leq \frac {(a + b + c)^2}{(ab + bc + ca)^2};
Which follows from the method in which I proved the previous inequality.
\boxed{24}
For positive reals x,y,z, prove that we always have the following inequality:
\sum_{cyc}\frac {x(y + z)^2}{2x + y + z} + \frac 92 \left(\frac {x + y + z}{xy + yz + zx}\right)^2\geq 6 + \frac 9{2(xy + yz ...
(Not Posted Anywhere Yet)
(To be updated as soon as I make another new problem)

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