TOP 10 problems of Week#3

 This week was full algebra biased!! 😄 

This is my first time trying inequalities, so this pure beginners level.  Do try all the problems first!! And if you guys get any nice solutions , do post in the comments section! 

These problem uses only Power mean Inequality  and Titu's lemma.

The First few problems happen to be  not problems, but tricks(?) which are extensively used.. 

Here are the walkthroughs of this week's top 10 Inequality problems!

10th position: Prove that for any real $a>0$ , $a+\frac{1}{a}\ge 2$

Walkthrough: a.  only AM-GM

9th position:Prove that for any real $a>0$ , $\frac{a}{1+a^2}\le \frac {a}{2a}$.

Walkthrough: a. Only AM-GM 

b. Use AM-GM to show that  $1+a^2\ge 2a $ 

8th position: Prove that for any real $x,y>0$ ,$\frac{1}{x+y}\le \frac{1}{4x}+\frac{1}{4y}$

Walkthrough: a. AM-HM inequality (cute 💖) 

7th position : Prove that for any real positive $p,q >0$ and $p+q=1$, then $\left(p+\frac{1}{p} \right)^2+ \left(q+\frac{1}{q} \right)^2 \ge \frac{25}{2}$

Walkthrough: a. Open the brackets , what do we get?

b. $p^2+\frac{1}{p^2}+2+q^2+\frac{1}{q^2}+2$  $ \ge \frac{25}{2}$

c. Note that by AM-GM, $p^2+q^2\le \left(\frac{p+q}{2}\right )^2 =\frac{1}{2}$

d. Again by AM-GM $\frac{1}{p^2}+\frac{1}{q^2}\ge \frac{2}{p\cdot q}$

e. hmm..still not achievable, so use this $(p+a)^2\ge 4pq$ which is true by AM-GM. 

f. conclude !

6th position: Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$  are different positive real numbers

Walkthrough: a. It's pure AM-GM

b. Note that by AM-GM $$\frac{ab}{c^3}+\frac{bc}{a^3}\ge \frac{2b}{ac}$$ 

c. Again by AM-GM $$\frac {b}{ac}+\frac{c}{ab}\ge \frac{2}{a}$$. 

5th position : Prove that for any real $a,b,c>0$ , $a^2+b^2+c^2\ge ab+bc+ca$.

Walkthrough: This is only AM-GM

a.  Note that $a^2+b^2\ge 2ab$ , sum stuffs and conclude!

4th Position:  Prove that if $a,b,c >0$ and $a^2+b^2+c^2=3 $ , then $\sum_{cyc}\frac{1}{1+ab}\ge \frac{3}{2}$

Walkthrough: Thankyou Dapper :) Do check out his blog in AOPS !!

a. So by Titu, $ \sum_{cyc} \frac {1^2}{1+ab}\ge \frac {3^2}{1+ab+1+bc+1+ca}= \frac{9}{3+ab+bc+ca}$

b. Use problem 5th and get that $\frac{9}{3+ab+bc+ca} \ge \frac{9}{6}=\frac{3}{2}$ .

3rd position : Prove that for any real $a+b\ge 1$ , then $a^4+b^4\ge \frac{1}{8}$

Walkthrough: a. Apply Quadratic Mean $\ge$ Arithmetic Mean 

I added it in the 3rd position , because I didn't know what Quadratic Mean was :P

2nd position : Let $x_0>x_1>x_2>\ldots>x_n$ be real numbers.

Prove $ x_{0}+\frac{1}{x_{0}-x_{1}}+\frac{1}{x_{1}-x_{2}}+\ldots+\frac{1}{x_{n-1}-x_{n}}\geq x_{n}+2n$.

Walkthrough: I love this guy's solution and couldn't resist to post walkthrough here, so full credits to him!

a.  Take $a_k=x_{k-1}-x_k$, simplify LHS.

b. Okie this is why problem 1 is so important , use the fact that $a+\frac{1}{a}\ge 2$ and you are done!! cute right?

1st Position:  Solve the system of equations in $\Bbb{R}^{+}$  

$$ a+b+c+d=12 $$ and $$abcd=27+ab+ac+ad+bc+bd+cd$$ 

Walkthrough: Oooo okie 4 variables 2 equations, seems unsolvable right? :P

a. Apply AM-GM in first equation and get $81\le abcd$.

b. Again apply AM-GM in the second equation (on $9+9+9+ab+ac+ad+bc+bd+cd$ ) , and get $abcd\ge 81$ .

c. Note that equality happens in AM-GM only when $a=b=c=d$.

d. Hence solution is $\boxed{a=b=c=d=3}$.

PS. : This YT video and series has so nice inequalities video, one can try it!!

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So these were my top 10 !

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! To follow ( if you want to ) click the 3 bars thing on the right.  See you all soon 😊.

Sunaina 💜