Top 10 Problems week#4 (it's late)

Okie!! Fine, I am late by 2 weeks but I was busy in OTIS submissions. And yayy!! I learnt how to use Evan.sty .  These problems are the exercises from a Titu handout in this website . Here's the full magazine , go to page 40's and one can find it :)

So full Titu :P

10th position (IMO shortlist, 1996): Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that $$\frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1.$$

Walkthrough: Thanku Rohan Bhaiya 😄

 a.   $$\sum_{cyc} \frac{ab}{a^5+b^5+ab}\le \sum_{cyc}\frac{c}{a+b+c}=1.$$

b. Cross Multiplying, it is enough to show that $$a^2b+ab^2+abc\le a^5c+b^5c+abc .$$

c. Multiply $abc=1$ to each side and use Muirhead.

9th position(RMO, 2006):If $a,b,c$ are three positive real numbers, prove that $$\frac {a^{2}+1}{b+c}+\frac {b^{2}+1}{c+a}+\frac {c^{2}+1}{a+b}\ge 3$$

Walkthrough: a. Using Titu, get  $$\frac {a^{2}+1}{b+c}+\frac {b^{2}+1}{c+a}+\frac {c^{2}+1}{a+b}\ge \frac{(a+b+c)^2 +9}{4(a+b+c)}+\frac{3}{2}$$

b. Use AM-GM, we get $$\frac{a+b+c}{4} + \frac{9}{4(a+b+c)}\ge \frac{3}{2}$$ .

8th position(RMO, 2012):  Given real numbers $a, b, c, d, e \ge 1$, prove that 

$$\frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge 20$$

Walkthrough: Same as prev problem :P

a. Use AM-GM, we get $$a+b+c+d+e-5 + \frac{25}{a+b+c+d+e-5}\ge 2\cdot 5=10$$ .

7th position(IMO, 1995) :  Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that

$$\frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.$$

Walkthrough: a. well this is well known :P, Note that $$\sum_{cyc} \frac{1}{a^3(b+c)}=\sum_{cyc} \frac{b^2c^2}{a(b+c)}.$$

b. apply titu and use  $ab+ca+bc\ge 3\sqrt[3]{ac\cdot bc\cdot ab}=3.$

6th Position(JMBO, 2003): Let $x, y, z > -1$. Prove that $$\frac{1+x^2}{1+y+z^2} + \frac{1+y^2}{1+z+x^2} + \frac{1+z^2}{1+x+y^2} \geq 2.$$

Walkthrough: a.Note that $x\le \frac{1+x^2}{2},$

b.apply titu and use $a=1+x^2, b=1+y^2, c=1+z^2$

c. Use  $$a^2+b^2+c^2\ge ab+bc+ca\Rightarrow (a+b+c)^2\ge 3(ab+bc+ca)\Rightarrow \frac{4(a+b+c)^2}{6(ab+bc+ca)}\ge. 2$$

5th position(Ireland, 1999): The sum of positive real numbers $a,b,c,d$ is $1$. Prove that:

$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+d}+\frac{d^2}{d+a} \ge \frac{1}{2}$$

Walkthrough: a. use titu directly :P

4th position(Moldova, 2007):  .Let $w, x, y, z$ be positive real numbers, prove that

$$\sum_{cyc}\frac{w}{x+2y+3z}\ge \frac{2}{3}$$

Walkthrough:Thanku NJOY 

a. Use Titu to get,  $$\sum_{cyc} \frac{w}{(x+2y+3z)} =\sum_{cyc} \frac{w^2}{(wx+2wy+3wz)}\ge \frac{(w+x+y+z)^2}{4(wx+xy+xz+wz+wy+yz)}.$$

b.Using AM-GM, Note that $$3(w^2+x^2+y^2+z^2)\ge  2( wx+xy+xz+wz+wy+yz)$$

3rd position(ADMO, 1991): Let $a_1 , a_2 , \dots a_n$ and $b_1 , b_2 , \dots b_n$ be positive numbers with $a_1 +a_2 +\dots+a_n = b_1 + b_2 + \dots+ b_n$ . Prove that

$$\frac{a_1^2}{a_1+b_1}+\dots+\frac{a_n^2}{a_n+b_n}\ge \frac{a_1 +a_2 +\dots+a_n}{2}$$

Walkthrough: a. this is direct :P

2nd position (St. Petersburg, 1999): Let $x_0>x_1>\dots >x_n$ be real numbers. Prove that $$x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n$$

Walkthrough: check out week 3 blog !! 

1st Position(Balkan, 1984): Let $n \geq 2$ be a positive integer and $a_{1},\ldots , a_{n}$ be positive real numbers such that $a_{1}+...+a_{n}= 1$. Prove that:

$$\frac{a_{1}}{1+a_{2}+\cdots +a_{n}}+\cdots +\frac{a_{n}}{1+a_{1}+a_{2}+\cdots +a_{n-1}}\geq \frac{n}{2n-1}$$

Walkthrough: a. Note that $$\frac{a_{1}}{1+a_{2}+\cdots +a_{n}}+\cdots +\frac{a_{n}}{1+a_{1}+a_{2}+\cdots +a_{n-1}}\ge \frac{a_{1}}{2-a_1}+\dots+\frac{a_n}{1-a_n}.$$

b. Note that $$\frac{a_1}{2-a_i}=-1+2\left(\frac{1^2}{2-a_i}\right) .$$

c. Use Titu :P

So these were my top 10 ! If u want to see the solutions , use this

What are your top 10s ? do write in the comments section (at least write something ! I will be happy to hear your comments ). Follow this blog if you want to see more contest math problems! See you all soon 😊.

Sunaina 💜