Analysis Quiz 16: Multiple-Choice Math (Square Root Inequality)

Question 1: Which of the methods below do you think can be used to prove $\sqrt{3}-\sqrt{2}\gt \sqrt{4}-\sqrt{3}$?

A. AM-GM Inequality.

B. Jensen's Inequality.

C. Cauchy–Schwarz inequality.

D. Squaring both sides of the equation and squaring again to remove the square root to make numerical comparison.

A collection of intriguing competition level problems for secondary school students.

Monday, December 28, 2015

Saturday, December 26, 2015

Wednesday, December 23, 2015

### Prove $\small a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$ for all real $a,\,b,\,c\ge 0$: Second Attempt

For reals $a,\,b,\,c\ge 0$, prove the inequality

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

Monday, December 21, 2015

### Prove $a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$ for all real $a,\,b,\,c\ge 0$: First Attempt

For reals $a,\,b,\,c\ge 0$, prove the inequality

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

$a(a-c)^2+b(b-c)^2\ge (a-c)(b-c)(a+b-c)$

and state when the equality holds.

Saturday, December 19, 2015

### Schur's inequality

I believe Schur's inequality might, perhaps, to the majority of the students that it's an inequality formula that students would not use much in their math problems.

But Schur's inequality is really a very magical and helpful inequality that if we are sufficiently familiarized with it for the special cases when $t=1$ and $t=2$, then we could use this "weapon" to tackle for lots of hard inequality IMO/competition problems.

But Schur's inequality is really a very magical and helpful inequality that if we are sufficiently familiarized with it for the special cases when $t=1$ and $t=2$, then we could use this "weapon" to tackle for lots of hard inequality IMO/competition problems.

Thursday, December 17, 2015

### Sophie Germain's Identity

I would wager not all of you know or sufficiently familiar with the identity as shown below:

$a^{4} + 4b^{4}=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$

It's actually a famous identity and it has a fancy name as well...it is called the Sophie Germain's Identity.

As the name suggests, Sophie Germain's identity was first discovered by Sophie Germain.

$a^{4} + 4b^{4}=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$

It's actually a famous identity and it has a fancy name as well...it is called the Sophie Germain's Identity.

As the name suggests, Sophie Germain's identity was first discovered by Sophie Germain.

Monday, December 7, 2015

### IMO Inequality problem

For the positive real numbers $x,\,y$ and $z$ that satisfy $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3$, prove that

$\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}\le \dfrac{3}{\sqrt{2}}$.

My solution:

$\dfrac{1}{\sqrt{x^3+1}}+\dfrac{1}{\sqrt{y^3+1}}+\dfrac{1}{\sqrt{z^3+1}}\le \dfrac{3}{\sqrt{2}}$.

My solution:

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