Serengeti22 wrote: Sorry this is not the best soulution. it isn't even a solution xD

Herr Schur in disguise. We have equivalently $(a+b+c)^2+\frac{3abc}{ab+bc+ca}\ge 4(ab+bc+ca)$ or $(a+b+c)^3+\frac{(a+b+c)^2}{ab+bc+ca}\cdot 3abc\ge 4(a+b+c)(ab+bc+ca)$ but $\frac{(a+b+c)^2}{ab+bc+ca}\ge 3$ hence proved.

DavyDuf wrote: Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\]

idea

Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. you gave me a good laugh man

We can also use SOS. Homogenize and it is equivalent to prove $\sum S_a(b-c)^2\geq 0$, $S_a=1-\dfrac{a^2}{ab+bc+ca}$. W.L.O.G $a\geq b\geq c$, then $S_a\leq S_b\leq S_c$. It's trivial that $S_b, S_c\geq 0$. Also $a^2S_b+b^2S_a=a^2+b^2-\dfrac{2a^2b^2}{ab+bc+ca}\geq 2ab-\dfrac{2a^2b^2}{ab}=0$.

melowmolly wrote: Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. you gave me a good laugh man I think maybe he is new to inequality. To prove an inequality, it's not only to verify when the equality may occur, but also prove it's true for any cases.

supercarry wrote: melowmolly wrote: Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. you gave me a good laugh man I think maybe he is new to inequality. To prove an inequality, it's not only to verify when the equality may occur, but also prove it's true for any cases. yeah middle schoolers are definitely new to inequalities

Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. Such a beautiful and elegant solution !!!!!!

megarnie wrote: Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. Such a beautiful and elegant solution !!!!!! just like me

Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution.

Serengeti22 wrote: Lets assume a,b,c are each 1/3 Therefore ((1/9)x3)+1/3 is greater than or equal to 2(3/9) So 2/3=2/3 This proves that this works If a,b,and c were not equal then the inquealtiely would still work just would not have been equal to. Sorry this is not the best soulution. bro solved world hunger

Slightly different proof. Let $ab+bc+ca=q$ and $abc=r$. Clearly $1=(a+b+c)^2\ge 3q$, so $q\le 1/3$. Adding $2q$ to both sides, it boils down proving $1+3r/q \ge 4q\iff 3r\ge q(4q-1)$. Note that there is nothing to prove if $q<1/4$, so assume $1/4\le q\le 1/3$. Schur's inequality yields $\sum a(a-b)(a-c)\ge 0$, that is $\sum a^3 + 6abc \ge (a+b+c)(ab+bc+ca)=q$, so that $\sum a^3 \ge q-6r$. Furthermore, using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, we get $\sum a^3 = 3r+1-3q$, so we arrive at $9r\ge 4q-1$. So, it suffices to verify $4q-1\ge 3q(4q-1)\iff (4q-1)(1-3q)\ge 0$, which holds precisely when $1/4\le q\le 1/3$.

The inequality is equivalent to $$(a^2+b^2+c^2)(ab+bc+ca)+3abc(a+b+c)\geq 2(ab+bc+ca)^2.$$If we expand things will cancel nicely to leave us with $$a^3b+ab^3+b^3c+bc^3+c^3a+ca^3\geq 2(a^2b^2+b^2c^2+c^2a^2),$$which is just $[3,1,0]\geq [2,2,0]$; true by Muirhead. $\blacksquare$

Let $a,b,c$ be three positive real numbers such that $1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge 1$$ DavyDuf wrote: Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\] Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Prove that$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4(ab+bc+ca) $$Let $a$, $b$ and $c$ be positive real numbers such that $abc=1.$ Prove that$$a^2+b^2+c^2 +\frac{9}{a+b+c} \geq 2(ab+bc+ca)$$

DavyDuf wrote: Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\]

Attachments:

This is my first posted solution to AOPS so Let us start first The request inequality can be written as $a^2 + b^2 + c^2 - 2ab - 2ac - 2ab>=- 3/1/a+1/b+1/c$ The inequality will be like $\sum{(a-b)^2}>= -3/1/a+1/b+1/c+ a^2 +b^2 +c^2 $ Notice that we want to prove that the RHS >=0 Now multi the RHS be $1/a + 1/b +1/c$ And we will get after subsitute a+b+c=1 The famous inequality $\sum{(a^2 +b^2)/c)}>=2$ And so we are done

Sorry for bad LATEX

Joider wrote: This is my first posted solution to AOPS so Let us start first The request inequality can be written as $a^2 + b^2 + c^2 - 2ab - 2ac - 2ab>=- 3/1/a+1/b+1/c$ The inequality will be like $\sum{(a-b)^2}>= -3/1/a+1/b+1/c+ a^2 +b^2 +c^2 $ Notice that we want to prove that the RHS >=0 Now multi the RHS by $1/a + 1/b +1/c$ And we will get after subsitute a+b+c=1 The famous inequality $\sum{(a^2 +b^2)/c)}>=2$ And so we are done

We need to prove that $a^{2}+b^{2}+c^{2}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq 2(ab+bc+ca).$ Since $a^{2}+b^{2}+c^{2}\geq ab+bc+ca,$ we just need to prove that $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq ab+bc+ca.$ From AM-HM inequality we get $\frac{1}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}.$ Now we just need to prove that $ \frac{1}{3}\geq ab+bc+ca.$ Since $a+b+c=1,$ $(a+b+c)^{2}=1=a^{2}+b^{2}+c^{2}+2(ab+bc+ca).$ From $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$ we get $\frac{1}{3}\geq ab+bc+ca.$ And we are done.(This is my first posted solution as well so i would be glad if you helped me improve my proof writing)