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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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! – Gurīdo Cuong · 10 months, 3 weeks ago

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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 ? – Gurīdo Cuong · 10 months, 2 weeks ago

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– Ms Ht · 10 months, 2 weeks ago

My solution have a problem and I am fixing it :DLog in to reply

Some of these are insane! – Calvin Lin Staff · 10 months, 2 weeks ago

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– Ms Ht · 10 months, 2 weeks ago

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.Log in to reply

– Gurīdo Cuong · 10 months, 2 weeks ago

The HSGS inequality is the hardest one yet to meLog in to reply

– Ms Ht · 10 months, 2 weeks ago

2 last inequality also.Log in to reply

– Ameya Daigavane · 10 months, 2 weeks ago

Vietnam TSTs are known for their difficulty!Log in to reply

– Ms Ht · 10 months, 2 weeks ago

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

– Akhash Raja Raam · 10 months, 2 weeks ago

I am really a person who needs help in maths and you have helped me earlier. I request the same again, please I need to learn from you MS HT ( Fact is I really don't know you but I know you have what I really need right now - BRILLIANCE!!) Please share me some of your knowledge!!!Log in to reply

we can solve problem 5 using titus lemma followed by am gm inequality right – Abhishek Alva · 10 months, 2 weeks ago

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– Gurīdo Cuong · 10 months, 2 weeks ago

Can you show me how you do itLog in to reply

– Abhishek Alva · 10 months, 2 weeks ago

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 latexLog in to reply

– Gurīdo Cuong · 10 months, 2 weeks ago

Ah yes, I've tried this approach but after that I found a mistake in it so I have to switch to another one. Here's why it's wrong. Titu's Lemma gives you \[LHS\geq\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\] But then due to \(3(a^2+b^2+c^2)\geq (a+b+c)^2\Rightarrow a^2+b^2+c^2\geq a+b+c\), we see that \(\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}\leq\frac{(a+b+c)^2}{2(a+b+c)}=RHS\), after all this process we'll still can't confirm whether \(LHS\geq RHS\) or not. So now most people make this misconception \(A\geq C, B\geq C \Rightarrow A\geq B\) and got their solution falseLog in to reply

– Abhishek Alva · 10 months, 2 weeks ago

thanksLog in to reply

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. – Akhash Raja Raam · 10 months, 2 weeks ago

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– Ms Ht · 10 months, 2 weeks ago

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

– Ms Ht · 10 months, 2 weeks ago

Ok I will help you as I can.ThanksLog in to reply

– Ms Ht · 10 months, 2 weeks ago

Thank you for your opinion. :DLog in to reply

– Gurīdo Cuong · 10 months, 2 weeks ago

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

– Ms Ht · 10 months, 2 weeks ago

Ok I will help!Log in to reply

– Akhash Raja Raam · 10 months, 2 weeks ago

I know that you are good enough for me to learn from you. Thanks for accepting my request!Log in to reply

– Ms Ht · 10 months, 2 weeks ago

No welcome,thank youLog in to reply

– Akhash Raja Raam · 10 months, 2 weeks ago

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

– Ms Ht · 10 months, 2 weeks ago

Account settings then you can change.Log in to reply

– Akhash Raja Raam · 10 months, 2 weeks ago

Thanks MS HT! And I want to discuss further using my gmail account. Its : akhashrajaraam@gmail.com . Its not restricted to anybody, I am ready to teach or learn from any of you!! I hope I didn't do something wrong here.Log in to reply