socrates wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 5.}$$ $\sum_{cyc}\frac{1+ab}{a+b}=3+\sum_{cyc}\frac{c+ab}{a+b}=3+\sum_{cyc}\frac{(c+a)(c+b)}{a+b}\geq3+\sum_{cyc}(a+b)=5$.

Very nice. Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{a+bc}{b+c}+\frac{b+ca}{c+a}+\frac{c+ab}{a+b}\geq 2.}$$

socrates wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 5.}$$ We have \[\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}-5=\frac{\sum{(a-b)^2(c-1)^2}}{2(a+b)(b+c)(c+a)}\ge{0}\]

sqing wrote: Very nice. Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{a+bc}{b+c}+\frac{b+ca}{c+a}+\frac{c+ab}{a+b}\geq 2.}$$ Just similar: $\frac{a+bc}{b+c}=\frac{(a+c)(a+b)}{b+c}$ And we have: $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\geq x+y+z$ obvious true since $(xy+yz+xz)^2 \geq 3xyz(x+y+z)$ So we done!

Yweh. Thanks.

the following let a,b,c >0 such that a+b+c=2 chứng minh $\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a} \geq 2$

NHN wrote: the following let a,b,c >0 such that a+b+c=2 chứng minh $\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a} \geq 2$ $\frac {a^2+bc}{b+c}+a=\frac {a^2+ab+bc+ca}{b+c}=\frac {(a+b)(c+a)}{b+c}$. Similarly, we have $\frac {b^2+ca}{c+a}=\frac {(b+c)(a+b)}{c+a}$ and $\frac {c^2+ab}{a+b}=\frac {(c+a)(b+c)}{a+b}$. Since $a+b+c=2$, the inequality returns to $\frac {(a+b)(c+a)}{b+c}+\frac {(b+c)(a+b)}{c+a}+\frac {(c+a)(b+c)}{a+b}\geq4.$. Using $\frac {xy}{z}+\frac {yz}{x}+\frac {zx}{y}\geq x+y+z$, we have $\frac {(a+b)(c+a)}{b+c}+\frac {(b+c)(a+b)}{c+a}+\frac {(c+a)(b+c)}{a+b}\geq (a+b)+(b+c)+(c+a)=2(a+b+c)=4$, as desired.

NHN wrote: the following let $a,b,c >0$ such that $a+b+c=2$ chứng minh $\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a} \geq 2$ The problem is equivalent to Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq 1.}$$

NHN wrote: the following let a,b,c >0 such that a+b+c=2 chứng minh $\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a} \geq 2$ We have \[\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a}-2=\frac{1}{2}\sum{\frac{(a+b)(a-b)^2}{(a+c)(b+c)}}\ge{0}\]

Nice. \[\frac {a^2+bc}{b+c}+\frac {b^2+ac}{a+c}+\frac {c^2+ba}{b+a}-(a+b+c)=\frac{1}{2}\sum{\frac{(a+b)(a-b)^2}{(a+c)(b+c)}}\]

socrates wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 5.}$$ Substract $3$ from each side then we get $\sum\frac{(1-a)(1-b)}{a+b}\geq2$ it is equivalent to $\sum\frac{(a+c)(b+c)}{a+b}\geq2$ let make substition $a+b=x,a+c=y$ and $b+c=z$ where $x+y+z=2$. We must prove $\sum\frac{xy}{z}\geq2$ $\frac{xy}{z}+\frac{zy}{x}\geq2y$ $\frac{xy}{z}+\frac{xz}{y}\geq2x$ $\frac{xz}{y}+\frac{zy}{x}\geq2z$ By Summing we get 2$\sum\frac{xy}{z}\geq4(x+y+z)$ $\to$ $\sum\frac{xy}{z}\geq2$

Let $ \displaystyle{a,b,c}$ be non-nagetive real numbers such that $\displaystyle{a+b+c=3.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 3.}$$

let a=1,2 b=0,8 c=1

socrates wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 5.}$$ Where did you find this problem? It was Serbia EGMO TST 2015 P2

Wolowizard wrote: socrates wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a+b+c=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 5.}$$ Where did you find this problem? It was Serbia EGMO TST 2015 P2 Peru TST 2010

Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 3.}$$

sqing wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 3.}$$ Try $a=1,b=\frac{1}{45},c=45$. But the following inequality is true! Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 2.}$$

szl6208 wrote: But the following inequality is true! Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}\geq 2.}$$ Equality holds when?

The equality does not hold, but the $2$ is a best estimation.

Thank arqady. Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}> 2.}$$

sqing wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{a+b}+\frac{1+bc}{b+c}+\frac{1+ca}{c+a}> 2.}$$ Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3.}$$here

sqing wrote: Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{abc=1.}$ Prove that $$ \displaystyle{\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3.}$$ If $abc=1,$ then $(1+ab)(1+bc)(1+ca)=(abc+ab)(abc+bc)(abc+ca)=a^2b^2c^2(1+a)(1+b)(1+c)=(1+a)(1+b)(1+c).$ So, by $\text{AM-GM inequality}$ we have: $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\geq 3\sqrt[3]{\frac{(1+ab)(1+bc)(1+ca)}{(1+a)(1+b)(1+c)}}=3,$ as desired.

Let $ \displaystyle{a,b,c}$ be positive real numbers such that $\displaystyle{a^2+b^2+c^2=3.}$ Prove that$$\frac{a+b}{1+ab}+\frac{b+c}{1+bc}+\frac{c+a}{1+ca}\ge 3.$$

Posting for storage Notice that if we subtract $3$ from both sides we obtain $\sum_{cyc}\frac{1+ab}{a+b}-1\ge2\Longleftrightarrow\sum_{cyc}\frac{(1-a)(1-b)}{a+b}\ge2$ However since $a+b+c=1$ the inequality can be rewritten as $\sum_{cyc}\frac{(b+c)(c+a)}{a+b}\ge2$, furthermore let $a+b=x, b+c=y\text{ and }c+a=z$ thus we obtain $x+y+z=2$ The inequality transforms into $\sum_{cyc}\frac{xy}{z}\ge2\Longleftrightarrow\frac{\sum_{cyc}x^2y^2}{xyz}\ge2\Longrightarrow\sum_{cyc}x^2y^2\ge2xyz=xyz\sum_{cyc}x=x^2yz+y^2zx+z^2xy$ Which is true by $\text{ Muirhead since }(2,2,0)\succ(2,1,1)$ $\blacksquare$.