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Monday, December 19, 2016

You can find ❤️ in ...... Math!


Saturday, December 17, 2016

Given the positive real numbers a,b,c and x,y,z satisfying the condition: a+x=b+y=c+z=1 Prove the inequality (abc+xyz)(1ay+1bz+1cx)3.

Given the positive real numbers a,b,c and x,y,z satisfying the condition:

a+x=b+y=c+z=1

Prove the inequality (abc+xyz)(1ay+1bz+1cx)3.

My solution:

Rewrite the intended LHS of the inequality strictly in terms of a, b and c, we have:

x=1a,y=1b,z=1c

(abc+xyz)(1ay+1bz+1cx)

=(abc+(1a)(1b)(1c))(1a(1b)+1b(1c)+1c(1a))

=(abc+1+ab+bc+ca(a+b+c)abc)(1a(1b)+1b(1c)+1c(1a))

=(1+ab+bc+ca(a+b+c))(1a(1b)+1b(1c)+1c(1a))

=(1a(1b)b(1c)c(1a))(1a(1b)+1b(1c)+1c(1a))

=1a(1b)+1b(1c)+1c(1a)1a(1b)(1b(1c)+1c(1a))

1b(1c)(1a(1b)+1c(1a))1c(1a)(1a(1b)+1b(1c))

=1b(1c)c(1a)a(1b)+1a(1b)c(1a)b(1c)+1a(1b)b(1c)c(1a)3

=1b+bcc+caa(1b)+1a+abc+acb(1c)+1a+abb+bcc(1a)3

=1bc(1b)+caa(1b)+1ca(1c)+abb(1c)+1ab(1a)+bcc(1a)3

=(1b)(1c)+caa(1b)+(1a)(1c)+abb(1c)+(1b)(1a)+bcc(1a)3

=1ca+c1b+1ab+a1c+1bc+b1a3

661cac1b1aba1c1bcb1a3 (By the AM-GM inequality, with 1a,1b,1c are all positive)

=63

=3 (Q.E.D.)


Thursday, December 15, 2016

Evaluate (2+3+6)(23+6)(2+36) without using a calculator.

Evaluate (2+3+6)(23+6)(2+36) without using a calculator.

My solution:

(2+3+6)(23+6)(2+36)

=(2+3+6)(23+6)(2+36)(236)(236)

=23(2+3+6)

By the Cauchy-Schwarz inequality, we have:

2+3+6<1+1+12+3+6=33

Hence 23(2+3+6)>2333.

From 528<529 we get, after taking the square root on both sides and rearranging:

4<2333



On the other hand,

From 50\gt 49, we get:

\sqrt{2}\gt \dfrac{7}{5}

From 12\gt 9, we get:

\sqrt{3}\gt \dfrac{3}{2}

From 6\gt 4, we get:

\sqrt{6}\gt 2

Adding them up gives:

\sqrt{2}+\sqrt{3}+\sqrt{6}\gt 4.9

\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\lt \dfrac{23}{4.9}=4.69.

We can conclude by now that \displaystyle \left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor=4.