For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that: $$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$
Problem
Source: Azerbaijan 2022 JBMO TST
Tags: inequalities, Algebra - Inequality, algebra, JBMO TST, Azerbaijan, Junior
sqing
04.07.2022 02:59
Iora wrote:
For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that:
$$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$
Attachments:
Iora
04.07.2022 03:02
any ideas? ( My solution was wrong)
sqing
04.07.2022 03:13
For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that: $$ \frac{a^2+b^2}{a^2+b^2+k}+\frac{b^2+c^2}{b^2+c^2+k}+\frac{c^2+a^2}{c^2+a^2+k} \geq \frac{6}{2+k} $$Where $k\in N^+.$
Iora
07.07.2022 02:48
let $x=a^2+b^2,y=b^2+c^2,z=c^2+a^2$. Then We have $x+y+z \ge 6$ and:
$$ \sum_{cyc} \frac{x}{x+k} = -\sum_{cyc} \frac{k}{x+k}+3$$Therefore the inequality is equivalent to:
$$ \sum_{cyc} \frac{1}{x+k} \le \frac{3}{2+k} $$which is is equivalent to:
$$(2+k)[(x+k)(y+k)+(y+k)(z+k)+(z+k)(x+k)] \le 3[(x+k)(y+k)(z+k)]$$or
$$ -6k^2+k^2(x+y+z)+2k(xy+yz+zx)-4k(y+x+z)-2(xy+yx+zx)+3xyz \ge 0$$or
$$-6k^2+ (x+y+z)(k^2-4k)+(xy+yz+zx)(2k-2) +3xyz\ge 0$$But noting that $(x+y+z) \ge 6$ we have :
$$ -24k+(xy+yz+zx)(2k-2)+3xyz \ge 0$$After that, letting $f(x,y,z) = -24k+(xy+yz+zx)(2k-2)+3xyz$ and use $UVW$ method or, $g(x,y,z)= x+y+z+t-3$, and since they are contunious, we can use lagrange multipliers ( I will not write all the details, just the system equation one) and find $(x-y)(1-2k-3z)=0$ ( by factoring out). And since $k \in \mathbb{N}$, we have $x=y$. From other equations, we can get $y=z$ hence $x=y=z$.The rest is obvious
And another solution for the original inequality (Someone showed me it, the solution is not mine)
After manupilating the original inequality to $ \sum_{cyc} \frac{1}{a^2+b^2+1} \le 1$, we can use C-S and expanding to finish it:
$$\sum_{cyc} \frac{1}{a^2+b^2+1}= \sum_{cyc} \frac{c^2+2}{((a^2+b^2+1)(1+1+c^2))} \le \sum_{cyc} \frac{c^2+2}{(a+b+c)^2}$$Which is trivially smaller than $1$.
This was probably the official solution too
sqing
07.07.2022 03:18
Let $a,b,c> 0 $ and $ \frac{1}{a^2}+ \frac{1}{b^2}+ \frac{1}{c^2}\leq \frac{\sqrt 3}{abc}.$ Prove that $$ab+bc+ca\leq 1$$https://artofproblemsolving.com/community/c6h1755159p28738441
mihaig
07.07.2022 05:06
Iora wrote: For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that: $$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$ Nice problem
sqing
10.07.2022 05:42
Iora wrote: For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that: $$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$ $$\iff$$Let $a,b,c>0$ and $ab+bc+ca\geq 3.$ Prove that $$(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq 2(a^2+b^2+c^2+1)$$
Attachments:
FriIzi
26.11.2022 20:16
Hello i think my solution shorter than yours Before the solution i dont now how to write solutions in latex forum and forgive me for that if you dont understand my solution tag me or reply me i will feedback for a short time 1/a + 1/b + 1/c >= 3/abc By that 1.~ab+bc+ca>=3~ If we look at the numerator we see that a² + b² >= (a+b)² / 2 We Demote the numerator and our inequality like that ~~SUM(cyc) (a+b)² / 2(a²+b²+1) >= 2~~ And if we look at the denominator we Enchance the denominator we do it like that ~~~SUM(cyc) (a+b)² /( 2(a+b)² + 2) >= 2~~~(Left hand side) So By Cauchy-Schwarz inequality special forum which we called Titu's Lemma in Azerbaijan Left hand side>= (2(a+b+c))²/(2(a²+b²+c²+3+ab+ac+bc)) >=2 And (a+b+c)²>=a²+b²+c²+3+ab+bc+ac And we see that we get ab+bc+ca>=3 And we solve the problem Good luck mates!!
Ferum_2710
27.04.2023 15:58
FriIzi wrote: Hello i think my solution shorter than yours Before the solution i dont now how to write solutions in latex forum and forgive me for that if you dont understand my solution tag me or reply me i will feedback for a short time 1/a + 1/b + 1/c >= 3/abc By that 1.~ab+bc+ca>=3~ If we look at the numerator we see that a² + b² >= (a+b)² / 2 We Demote the numerator and our inequality like that ~~SUM(cyc) (a+b)² / 2(a²+b²+1) >= 2~~ And if we look at the denominator we Enchance the denominator we do it like that ~~~SUM(cyc) (a+b)² /( 2(a+b)² + 2) >= 2~~~(Left hand side) So By Cauchy-Schwarz inequality special forum which we called Titu's Lemma in Azerbaijan Left hand side>= (2(a+b+c))²/(2(a²+b²+c²+3+ab+ac+bc)) >=2 And (a+b+c)²>=a²+b²+c²+3+ab+bc+ac And we see that we get ab+bc+ca>=3 And we solve the problem Good luck mates!! My guy how do you get $(a+b+c)^2 \geq a^2 + b^2 + c^2 + 3 + ab + ac + bc$
MAriOsig
27.04.2023 16:22
Ferum_2710 wrote: FriIzi wrote: Hello i think my solution shorter than yours Before the solution i dont now how to write solutions in latex forum and forgive me for that if you dont understand my solution tag me or reply me i will feedback for a short time 1/a + 1/b + 1/c >= 3/abc By that 1.~ab+bc+ca>=3~ If we look at the numerator we see that a² + b² >= (a+b)² / 2 We Demote the numerator and our inequality like that ~~SUM(cyc) (a+b)² / 2(a²+b²+1) >= 2~~ And if we look at the denominator we Enchance the denominator we do it like that ~~~SUM(cyc) (a+b)² /( 2(a+b)² + 2) >= 2~~~(Left hand side) So By Cauchy-Schwarz inequality special forum which we called Titu's Lemma in Azerbaijan Left hand side>= (2(a+b+c))²/(2(a²+b²+c²+3+ab+ac+bc)) >=2 And (a+b+c)²>=a²+b²+c²+3+ab+bc+ac And we see that we get ab+bc+ca>=3 And we solve the problem Good luck mates!! My guy how do you get $(a+b+c)^2 \geq a^2 + b^2 + c^2 + 3 + ab + ac + bc$ $$(a+b+c)^2=(a^2+b^2+c^2+ab+bc+ca)+(ab+bc+ca)$$from the problem $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ $\leftrightarrow$ $ab+bc+ca\geq3$ $$\therefore (a+b+c)^2\geq(a^2+b^2+c^2+ab+bc+ca)+3$$
Professor33
21.09.2023 10:33
sqing wrote:
Iora wrote:
For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that:
$$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$
How u got (a^2+b^2)(b^2+c^2)(c^2+a^2)>=8/9(a^2+b^2+c^2)(a^2.b^2+b^2.c^2+c^2,a^2)
zaidova
14.11.2024 23:08
Jensen kills this problem
anhduy98
15.11.2024 01:25
Iora wrote: For positive real numbers $a,b,c$, $\frac{1}{a}+\frac{1}{b} + \frac{1}{c} \ge \frac{3}{abc}$ is true. Prove that: $$ \frac{a^2+b^2}{a^2+b^2+1}+\frac{b^2+c^2}{b^2+c^2+1}+\frac{c^2+a^2}{c^2+a^2+1} \ge 2$$ $$LHS=3-\sum \frac{1+1+c^2}{(a^2+b^2+1)(1+1+c^2)}\ge 3-\frac{a^2+b^2+c^2+6}{(a+b+c)^2}\ge 3-\frac{a^2+b^2+c^2+2(ab+bc+ca)}{(a+b+c)^2}=2.$$ .