rightways wrote: $a,b,c$ are positive real numbers such that $a+b+c\ge 3$ and $a^2+b^2+c^2=2abc+1$. Prove that $$a+b+c\le 2\sqrt{abc}+1$$ 2023? Thanks. Let $a,b,c$ be nonnegative real numbers such that $a^2+b^2+c^2=3.$ Prove that $$a+b+c\geq 1+2\sqrt{abc}.$$

Yes, I will edit

rightways wrote: Yes, I will edit Thanks.

$2 \sqrt{abc}+1 \geq a+b+c$ $4abc \geq (a+b+c-1)^2$ $4abc \geq a^2+b^2+c^2+1+2ab+2bc+2ca-2a-2b-2c$ $4abc \geq 2abc+1+1+2ab+2bc+2ca-2a-2b-2c$ $abc \geq 1+ab+bc+ca-a-b-c$ $0 \geq (1-a)(1-b)(1-c) ...(1)$ Let's prove the last inequality. $1-a=x,1-b=y,1-c=z$ We have $x+y+z<0$ By rearrangement inequality $3xyz<x^3+y^3+z^3$ We'll prove that $x^3+y^3+z^3<0$ $2x^3+2y^3+2z^3<0$ $\sum (x+y)(x^2-xy+y^2)<0$ $x+y<-z$ so $\sum (x+y)(x^2-xy+y^2)<\sum -z(x^2-xy+y^2)$ $\sum -z(x^2-xy+y^2)<0$ $x^2y+x^2z+y^2x+y^2z+z^2x+z^2x>6xyz$ If $xyz \leq 0$, $(1)$ is obviously true. If $xyz>0$, $x^2y+x^2z+y^2x+y^2z+z^2x+z^2x>6xyz>3xyz$ Contradiction $\implies xyz \leq 0$

$(a+b+c-1)^2\leq 4abc$$\Rightarrow$ $4abc+2(a+b+c)\geq 2(ab+bc+ca)+1+a^2+b^2+c^2$ $abc+a+b+c\geq ab+bc+ca +1$$\Rightarrow$ $(a-1)(b-1)(c-1)\geq 0 $ lets prove all numbers $\geq 1$ or two of them $<1 $ Let all but we have $a+b+c>3$ Lets only one $<1$ let is $a$ $\Rightarrow$ $(a^2-1)(b^2-1)=(ab-c)^2\geq 0 $ $\Rightarrow$ $a,b<1$ or $a,b,c>1$Q.E.D.

KOMKZ wrote: $(a+b+c-1)^2\leq 4abc$$\Rightarrow$ $4abc+2(a+b+c)\geq 2(ab+bc+ca)+1+a^2+b^2+c^2$ $abc+a+b+c\geq ab+bc+ca +1$$\Rightarrow$ $(a-1)(b-1)(c-1)\geq 0 $ lets prove all numbers $\geq 1$ or two of them $<1 $ Let all but we have $a+b+c>3$ Lets only one $<1$ let is $a$ $\Rightarrow$ $(a^2-1)(b^2-1)=(ab-c)^2\geq 0 $ $\Rightarrow$ $a,b<1$ or $a,b,c>1$Q.E.D. Fantastic english mate

bruh how is this grade 9?

MetaphysicalWukong wrote: bruh how is this grade 9? and @KOMKZ how did you get the third line?

Let assume $a<1$.Then we have $(a^2-1)(b^2-1)=(ab-c)^2>=0$ from this we get $b<=1$ similarly $c<=1$.And we get $a+b+c<3$ but this is contraduction.Then $a,b,c>=1$ (*) After that $a+b+c-1<=2\sqrt{abc}$ $(a+b+c-1)^2<=4abc$ $2abc+2+2(ab+bc+ca)-2(a+b+c)<=4abc$ $1+ab+bc+ca-a-b-c<=abc$ $0<=(a-1)(b-1)(c-1)$ That is true by (*)