Let $a,b,c$ be positive reals such that $abc=1$. Prove the inequality $$\frac{4a-1}{(2b+1)^2} + \frac{4b-1}{(2c+1)^2} + \frac{4c-1}{(2a+1)^2}\geqslant 1.$$
Problem
Source: 2019 Thailand Mathematical Olympiad P5
Tags: algebra, inequalities, 2019
MarkBcc168
22.05.2019 14:00
Add one to each term and divide by $4$. This is equivalent to
$$\sum_{\mathrm{cyc}}\frac{b^2+b+a}{(2b+1)^2}\geqslant 1$$Now we can use Cauchy Schwarz in form $(b^2+b+a)(1+b+\frac{1}{a})\geqslant (b+b+1)^2$. Thus it suffices to prove that
$$\sum_{\mathrm{cyc}}\frac{1}{b+\frac{1}{a}+1}\geqslant 1.$$In fact, it turns out to be an equality. The cleanest way to verify that is to substitute $a=\frac{x}{y}, b=\frac{z}{x}, c=\frac{y}{z}$ and see that each term is equal to $\frac{x}{x+y+z}$.
MrOrange
22.05.2019 14:27
Also posted here.
sqing
27.05.2019 09:35
Let $a,b,c$ be positive reals such that $abc=1$. Prove the inequality $$\frac{ka-1}{((k-2)b+1)^2} + \frac{kb-1}{((k-2)c+1)^2} + \frac{kc-1}{((k-2)a+1)^2}\geqslant \frac{3}{k-1}.$$Where $2\leq k\leq 4 .$ MarkBcc168 wrote: Let $a,b,c$ be positive reals such that $abc=1$. Prove the inequality $$\frac{4a-1}{(2b+1)^2} + \frac{4b-1}{(2c+1)^2} + \frac{4c-1}{(2a+1)^2}\geqslant 1.$$
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sqing
28.11.2019 05:58
2018 Thailand Mathematical Olympiad Q4: Let $a,b,c$ be non-zero real number such that $a+b+c=0.$ Find the maximum of $\frac{a^2b^2c^2}{(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)}.$
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https://artofproblemsolving.com/community/c6h19230p128903
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sqing
02.11.2020 13:50
2020 Thailand Mathematical Olympiad Q2: Let $a,b,c$ be positive real number such that $a^2+b^2+c^2=3.$ Prove that$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq\frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$p/7043073437
Krits
28.12.2020 07:13
sqing wrote: 2020 Thailand Mathematical Olympiad Q2: Let $a,b,c$ be positive real number such that $a^2+b^2+c^2=3.$ Prove that$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq\frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$p/7043073437 FYI That's not from Thailand mathematical olympiad 2020, the only inequality problem appeared in the contest is as follow: $a,b,c\in \mathbb{R}^+ , a+b+c=3$, Prove that $$\frac{a^6}{c^2+2b^3}+\frac{b^6}{a^2+2c^3}+\frac{c^6}{b^2+2a^3}\geqslant 1.$$
sqing
28.12.2020 09:46
Krits wrote:
sqing wrote:
2020 Thailand Mathematical Olympiad Q2:
Let $a,b,c$ be positive real number such that $a^2+b^2+c^2=3.$ Prove that$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq\frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$p/7043073437
FYI That's not from Thailand mathematical olympiad 2020, the only inequality problem appeared in the contest is as follow:
$a,b,c\in \mathbb{R}^+ , a+b+c=3$, Prove that $$\frac{a^6}{c^2+2b^3}+\frac{b^6}{a^2+2c^3}+\frac{c^6}{b^2+2a^3}\geqslant 1.$$
2020 Thailand Mathematical Olympiad :
$\frac{a^6}{c^2+2b^3} + \frac{b^6}{a^2+2c^3} + \frac{c^6}{b^2+2a^3} \geq \frac{(a^3+b^3+c^3)^2}{2(a^3+b^3+c^3)+(a^2+b^2+c^2)}$ so we need to show $(a^3+b^3+c^3)^2\geq 2(a^3+b^3+c^3)+(a^2+b^2+c^2)$
$3(a^2+b^2+c^2)\geq (a+b+c)^2=9 \Rightarrow a^2+b^2+c^2\geq 3$
$9(a^3+b^3+c^3)\geq (a+b+c)^3=27 \Rightarrow a^3+b^3+c^3\geq 3$
$(a^3+b^3+c^3)^2=\frac{2(a^3+b^3+c^3)^2}{3}+\frac{(a^3+b^3+c^3)^2}{3}\geq 2(a^3+b^3+c^3)+\frac{(a^2+b^2+c^2)^3}{9}\geq 2(a^3+b^3+c^3)+(a^2+b^2+c^2)$
Krits
30.12.2020 14:17
sqing wrote: 2020 Thailand Mathematical Olympiad Q2: Let $a,b,c$ be positive real number such that $a^2+b^2+c^2=3.$ Prove that$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq\frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$p/7043073437 This problem is from Thailand TSTST camp 2020. Solution: One can prove by fully expanding and using the condition $a^2+b^2+c^2=3$ to get $\frac{c}{c+a+1}+\frac{a}{a+b+1}+\frac{b}{b+c+1}\leq 1$ which is equivalent to LHS$\geq 2$. Now RHS$\leq 2$ is true from c-s:$$(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc=\sum_{cyc}(\frac{a^2b^2}{a}+\frac{a^2b^2}{b})+\frac{4abc}{2}\geq (\frac{2}{a+b+c+1})(ab+bc+ca+\sqrt{abc})^2\geq \frac{(ab+bc+ca+\sqrt{abc})^2}{2}.$$
sqing
30.12.2020 15:43
Krits wrote:
sqing wrote:
2020 Thailand Mathematical Olympiad Q2:
Let $a,b,c$ be positive real number such that $a^2+b^2+c^2=3.$ Prove that$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq\frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$p/7043073437
This problem is from Thailand TSTST camp 2020.
Solution:
One can prove by fully expanding and using the condition $a^2+b^2+c^2=3$ to get $\frac{c}{c+a+1}+\frac{a}{a+b+1}+\frac{b}{b+c+1}\leq 1$ which is equivalent to LHS$\geq 2$.
Now RHS$\leq 2$ is true from c-s:$$(a+b)(b+c)(c+a)=\sum_{sym}a^2b+2abc=\sum_{cyc}(\frac{a^2b^2}{a}+\frac{a^2b^2}{b})+\frac{4abc}{2}\geq (\frac{2}{a+b+c+1})(ab+bc+ca+\sqrt{abc})^2\geq \frac{(ab+bc+ca+\sqrt{abc})^2}{2}.$$
Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3.$ Then $$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1\iff \frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq 2$$$$\frac{a+1}{c+a+1}+\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}\geq2\geq \frac{(ab+bc+ca+\sqrt{abc})^2}{(a+b)(b+c)(c+a)}.$$
sqing
30.12.2020 15:49
Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3.$ Prove that $$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1.$$
Krits
30.12.2020 16:13
sqing wrote: Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3.$ Prove that $$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1.$$ from fully expanding and substituting $x^2+y^2+z^2=3$, It's equivalent to $3xyz+2(x^2y+y^2z+z^2x)\leq x^3+y^3+z^3+xy^2+yz^2+zx^2+3$ which is true from $xyz\leq 1$ and $x^3+xy^2\geq 2x^2y$ and so on..
sqing
31.12.2020 03:25
Let $a,b,c$ be positive real numbers such that $abc=1.$ Prove that $$\frac{11}{5}>\frac{1}{3a^2-a+1}+\frac{1}{3b^2-b+1}+\frac{1}{3c^2-c+1} \geq 1$$$$\frac{67}{25}>\frac{1}{4a^2-2a+1}+\frac{1}{4b^2-2b+1}+\frac{1}{4c^2-2c+1}\geq 1$$h
Krits wrote:
sqing wrote:
Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3.$ Prove that
$$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1.$$
from fully expanding and substituting $x^2+y^2+z^2=3$, It's equivalent to
$3xyz+2(x^2y+y^2z+z^2x)\leq x^3+y^3+z^3+xy^2+yz^2+zx^2+3$
which is true from $xyz\leq 1$ and $x^3+xy^2\geq 2x^2y$ and so on..
$$\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1\iff 3abc+2(a^2b+b^2c+c^2a)\leq a^3+b^3+c^3+ab^2+bc^2+ca^2+3$$$$\iff3(1-abc)+a(a-b)^2+b(b-c)^2+c(c-a)^2\geq 0$$