A collection of intriguing competition level problems for secondary school students.
Monday, December 19, 2016
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=1−a,y=1−b,z=1−c
(abc+xyz)(1ay+1bz+1cx)
=(abc+(1−a)(1−b)(1−c))(1a(1−b)+1b(1−c)+1c(1−a))
=(abc+1+ab+bc+ca−(a+b+c)−abc)(1a(1−b)+1b(1−c)+1c(1−a))
=(1+ab+bc+ca−(a+b+c))(1a(1−b)+1b(1−c)+1c(1−a))
=(1−a(1−b)−b(1−c)−c(1−a))(1a(1−b)+1b(1−c)+1c(1−a))
=1a(1−b)+1b(1−c)+1c(1−a)−1−a(1−b)(1b(1−c)+1c(1−a))
−1−b(1−c)(1a(1−b)+1c(1−a))−1−c(1−a)(1a(1−b)+1b(1−c))
=1−b(1−c)−c(1−a)a(1−b)+1−a(1−b)−c(1−a)b(1−c)+1−a(1−b)−b(1−c)c(1−a)−3
=1−b+bc−c+caa(1−b)+1−a+ab−c+acb(1−c)+1−a+ab−b+bcc(1−a)−3
=1−b−c(1−b)+caa(1−b)+1−c−a(1−c)+abb(1−c)+1−a−b(1−a)+bcc(1−a)−3
=(1−b)(1−c)+caa(1−b)+(1−a)(1−c)+abb(1−c)+(1−b)(1−a)+bcc(1−a)−3
=1−ca+c1−b+1−ab+a1−c+1−bc+b1−a−3
≥66√1−ca⋅c1−b⋅1−ab⋅a1−c⋅1−bc⋅b1−a−3 (By the AM-GM inequality, with 1−a,1−b,1−c are all positive)
=6−3
=3 (Q.E.D.)
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=1−a,y=1−b,z=1−c
(abc+xyz)(1ay+1bz+1cx)
=(abc+(1−a)(1−b)(1−c))(1a(1−b)+1b(1−c)+1c(1−a))
=(abc+1+ab+bc+ca−(a+b+c)−abc)(1a(1−b)+1b(1−c)+1c(1−a))
=(1+ab+bc+ca−(a+b+c))(1a(1−b)+1b(1−c)+1c(1−a))
=(1−a(1−b)−b(1−c)−c(1−a))(1a(1−b)+1b(1−c)+1c(1−a))
=1a(1−b)+1b(1−c)+1c(1−a)−1−a(1−b)(1b(1−c)+1c(1−a))
−1−b(1−c)(1a(1−b)+1c(1−a))−1−c(1−a)(1a(1−b)+1b(1−c))
=1−b(1−c)−c(1−a)a(1−b)+1−a(1−b)−c(1−a)b(1−c)+1−a(1−b)−b(1−c)c(1−a)−3
=1−b+bc−c+caa(1−b)+1−a+ab−c+acb(1−c)+1−a+ab−b+bcc(1−a)−3
=1−b−c(1−b)+caa(1−b)+1−c−a(1−c)+abb(1−c)+1−a−b(1−a)+bcc(1−a)−3
=(1−b)(1−c)+caa(1−b)+(1−a)(1−c)+abb(1−c)+(1−b)(1−a)+bcc(1−a)−3
=1−ca+c1−b+1−ab+a1−c+1−bc+b1−a−3
≥66√1−ca⋅c1−b⋅1−ab⋅a1−c⋅1−bc⋅b1−a−3 (By the AM-GM inequality, with 1−a,1−b,1−c are all positive)
=6−3
=3 (Q.E.D.)
Thursday, December 15, 2016
Evaluate ⌊(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)⌋ without using a calculator.
Evaluate ⌊(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)⌋ without using a calculator.
My solution:
(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)
=(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)(−√2−√3−√6)(−√2−√3−√6)
=23(√2+√3+√6)
By the Cauchy-Schwarz inequality, we have:
√2+√3+√6<√1+1+1√2+3+6=√33
Hence 23(√2+√3+√6)>23√33.
From 528<529 we get, after taking the square root on both sides and rearranging:
4<23√33
∴
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.
My solution:
(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)
=(−√2+√3+√6)(√2−√3+√6)(√2+√3−√6)(−√2−√3−√6)(−√2−√3−√6)
=23(√2+√3+√6)
By the Cauchy-Schwarz inequality, we have:
√2+√3+√6<√1+1+1√2+3+6=√33
Hence 23(√2+√3+√6)>23√33.
From 528<529 we get, after taking the square root on both sides and rearranging:
4<23√33
∴
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.
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