Inequality for grade 10 mathematics gifted student (Hanoi,Viet Nam)

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

#Algebra

Easy Math Editor

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Here'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!

P C - 5 years ago

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 ?

P C - 5 years ago

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

Son Nguyen - 5 years ago

Some of these are insane!

Calvin Lin Staff - 5 years 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.

Son Nguyen - 5 years ago

Vietnam TSTs are known for their difficulty!

Ameya Daigavane - 5 years ago

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

Son Nguyen - 5 years ago

@Son Nguyen – 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!!!

Akhash Raja Raam - 5 years ago

The HSGS inequality is the hardest one yet to me

P C - 5 years ago

2 last inequality also.

Son Nguyen - 5 years ago

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 - 5 years ago

Thank you for your opinion. :D

Son Nguyen - 5 years ago

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

P C - 5 years ago

@P C – Ok I will help!

Son Nguyen - 5 years ago

@Son Nguyen – I know that you are good enough for me to learn from you. Thanks for accepting my request!

Akhash Raja Raam - 5 years ago

@Akhash Raja Raam – No welcome,thank you

Son Nguyen - 5 years ago

Ok I will help you as I can.Thanks

Son Nguyen - 5 years ago

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

Son Nguyen - 5 years ago

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

Akhash Raja Raam - 5 years ago

Account settings then you can change.

Son Nguyen - 5 years ago

@Son Nguyen – Thanks MS HT! And I want to discuss further using my gmail account. Its : [email protected] . 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.

Akhash Raja Raam - 5 years ago

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

abhishek alva - 5 years ago

Can you show me how you do it

P C - 5 years 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 latex

abhishek alva - 5 years ago

@Abhishek Alva – 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 false

P C - 5 years ago

@P C – thanks

abhishek alva - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$