I have some cool problem for us:You can read this

Problem 1:Taken from HSGS school:Problem

Problem 2 :Taken from Hanoi Pedagogic Highschool grade 10 selection test. Problem

Problem 3: Taken from Foreign Language Specialized School :Problem

Problem 4: Taken from Hanoi selection test grade 10 :Problem.

Problem 5: Taken from Hanoi Amsterdam High school; Chu Van An high school;Nguyen Hue High school selection test for mathematics gifted student.

Given that \(a;b;c\) are positive reals that satisfy \(a^2+b^2+c^2=3\).

Prove that \(\frac{a^2}{a+b^2}+\frac{b^2}{b+c^2}+\frac{c^2}{c+a^2}\geq \frac{a+b+c}{2}\)

Problem 6:Taken from Hanoi Amsterdam High school; Chu Van An high school;Nguyen Hue High school selection test for mathematics computer science gifted student.

Given \(a;b;c\) are positive reals number and \(a+b+c=1\)

Prove that \(\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}\leq \frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)

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## Comments

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TopNewestHere's my solution for problem 6. First, we set \(a=\frac{x}{3}, b=\frac{y}{3},c=\frac{z}{3}\), then the condition becomes \(x+y+z=3\) and the problem becomes \[\displaystyle\sum_{cyc}\frac{3x}{3x+y^2}\leq\frac{3}{4}\displaystyle\sum_{cyc}\frac{1}{x}\] \[\Rightarrow\displaystyle\sum_{cyc}\frac{4x}{3x+y^2}\leq\displaystyle\sum_{cyc}\frac{1}{x}\] Now, consider the \(LHS\) \[\displaystyle\sum_{cyc}\frac{4x}{3x+y^2}\stackrel{AM-GM}\leq\displaystyle\sum_{cyc}\frac{4x}{4\sqrt[4]{x^3}\sqrt{y}}=\displaystyle\sum_{cyc}\frac{\sqrt[4]{x}}{\sqrt{y}}\] By Cauchy-Schwarz \[\bigg(\displaystyle\sum_{cyc}\frac{\sqrt[4]{x}}{\sqrt{y}}\bigg)^2\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})\bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)\] We have these inequality \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\stackrel{Titu's}\geq\frac{9}{x+y+z}=3\) and \((\sqrt{x}+\sqrt{y}+\sqrt{z})^2\stackrel{C-S}\leq 3(x+y+z)=9\Rightarrow \sqrt{x}+\sqrt{y}+\sqrt{z}\leq 3\) \[\therefore \sqrt{x}+\sqrt{y}+\sqrt{z}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\] So now \((\sqrt{x}+\sqrt{y}+\sqrt{z})\bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)\leq\bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)^2\) \[\therefore \displaystyle\sum_{cyc}\frac{\sqrt[4]{x}}{\sqrt{y}}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\] Finally \[LHS=\displaystyle\sum_{cyc}\frac{4x}{3x+y^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=RHS\] The inequality is proven, equality holds when \(x=y=z=1\) or \(a=b=c=\frac{1}{3}\). @MS HT check it out!

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Solution for problem 5. Rewrite the \(LHS\) \[2(a+b+c)-\bigg(\frac{2ab^2}{a+b^2}+\frac{2bc^2}{b+c^2}+\frac{2ca^2}{c+a^2}\bigg)\] By AM-GM \[LHS\geq 2(a+b+c)-\bigg(\frac{2ab^2}{2b\sqrt{a}}+\frac{2bc^2}{2c\sqrt{b}}+\frac{2ca^2}{2a\sqrt{c}}\bigg)=2(a+b+c)-(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})\] By C-S inequality \[(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2\leq (a+b+c)(ab+bc+ac)\] We have these inequality \((a+b+c)^2\leq 3(a^2+b^2+c^2)\Rightarrow a+b+c\leq 3\) and \(ab+bc+ac\leq\frac{(a+b+c)^2}{3}\), all can be proven by C-S \[\therefore b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\leq a+b+c\] \[\Rightarrow LHS\geq 2(a+b+c)-(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})\geq a+b+c=RHS\] The inequality is proven, the equality holds when \(a=b=c=1\). @MS HT, can you show me your U.C.T approach ?

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My solution have a problem and I am fixing it :D

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Some of these are insane!

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Yes,I agree with you.My opinion is not agree with this type of our selection test from Hanoi Education.Some of them are nearly Olympic Test for high school student.I lost more than 1 hour to have a solution.

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Vietnam TSTs are known for their difficulty!

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Yes,so difficult. These problem are nearly to Olympic for grade 12.

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The HSGS inequality is the hardest one yet to me

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2 last inequality also.

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Though its shameful to say it publicly, the truth is that I couldn't even understand these questions even though I am studying in a grade a level higher than the ones who wrote this exam...Education in where I live is all money and no quality. I have very high dreams that education where I live isn't enough. That's why I approached you MS HT because according to your profile we are of same age. I felt that it would be better if I could learn from someone from my age, not too old and not too young. I hope I can get a good reply here in my comment for my request to learn from you.

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Thank you for your opinion. :D

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Dude, he wants you to be his instructor and help him in solving problems, at least don't let him down by saying that

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Ok I will help you as I can.Thanks

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But I'm not good as you think so I will help you as I can.Thank you!

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And can anybody also tell me how to change profile picture?

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Account settings then you can change.

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we can solve problem 5 using titus lemma followed by am gm inequality right

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Can you show me how you do it

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we write a^2+b^2+c^2 as a^2/1+b^2/1+c^2/1 by the titus lemma,[ (a+b+c)^2]/3=3 after solving the linear equation we get a+b+c=3 LHS- again by using the titus lemma we get- [ ( a+b+c)^2]/(a+b+c+a^2+b^2+c^2) we know that a+b+c=3 &a^2+b^2+c^2 by inserting the values and solving for the equation we get 3/2>or=3/2 hence proved. and by the way i dont know to use latex

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