Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\]
Problem
Source: Indonesian TST for IMO 2023 Stage 1: Test 2 - Algebra
Tags: algebra, inequalities proposed, Indonesia, Indonesia TST
28.02.2023 05:14
Serengeti22 wrote: Sorry this is not the best soulution. it isn't even a solution xD
28.02.2023 05:41
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.
28.02.2023 05:41
DavyDuf wrote: Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\]
28.02.2023 05:50
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
28.02.2023 05:51
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 $a^2S_b+b^2S_a=a^2+b^2-\dfrac{2a^2b^2}{ab+bc+ca}\geq 2ab-\dfrac{2a^2b^2}{ab}=0$.
28.02.2023 05:55
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.
28.02.2023 06:01
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
28.02.2023 06:37
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 !!!!!!
28.02.2023 06:39
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
28.02.2023 07:00
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.
28.02.2023 08:43
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
01.03.2023 01:59
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 $1/4\le q\le 1/3$.
01.03.2023 09:07
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. $\blacksquare$
01.03.2023 11:04
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$$a^2+b^2+c^2 +\frac{9}{a+b+c} \geq 2(ab+bc+ca)$$
03.03.2023 04:16
DavyDuf wrote: Let $a,b,c$ positive real numbers and $a+b+c = 1$. Prove that \[a^2 + b^2 + c^2 + \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \ge 2(ab + bc + ac)\]
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19.08.2023 21:00
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
19.08.2023 21:07
Sorry for bad LATEX
19.08.2023 21:08
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
20.08.2023 00:22
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)