Olympiad for grade 10

Algebra Level 4

Let $a,b,c> 0,abc=1$ Let the minimum of P is M $P= \frac{a^2}{\sqrt{2+2ab}}+\frac{b^2}{\sqrt{2+2bc}}+\frac{c^2}{\sqrt{2+2ca}}$ If $M=\frac{m}{n}$ Find the value of m+n if $\gcd{m,n}=1$

The answer is 5.

1 solution

Son Nguyen
Jan 16, 2016

By applying Titu's Lemma(Engel) we have: $P\geq \frac{(a+b+c)^2}{\sqrt{2+2ab}+\sqrt{2+2bc}+\sqrt{2+2ca}}$ Let $\sqrt{2+2ab}+\sqrt{2+2bc}+\sqrt{2+2ca}=N$ By applying C-S inequality: $N^2\leq 3(2ab+2bc+2ca+6)$ $2ab+2bc+2ca\leq 6\leq 2a^2+2b^2+2c^2(abc=1\rightarrow a^2+b^2+c^2\geq 3)$ $\Rightarrow N\leq 6$ Since $a+b+c\geq 3(abc=1)$ So $P\geq \frac{3}{2}=M=\frac{m}{n}$ $\rightarrow m+n=5$

Because $ab+bc+ca\leq \frac{(a+b+c)^{2}}{3}$

So $2ab+2bc+2ca\leq 6\leq 2a^2+2b^2+2c^2(abc=1\rightarrow a^2+b^2+c^2\geq 3)$ You can check back @Gurīdo Cuong

Son Nguyen - 5 years, 4 months ago

Rewrite N $N=\sqrt{2(1+ab)}+\sqrt{2(1+ac)}+\sqrt{2(1+bc)}$ Using Cauchy - Schwartz we get $N\leq\sqrt{6(3+ab+bc+ca)}$ We've got $9\leq(a+c+b)^2$ and $3(ab+bc+ac)\leq(a+b+c)^2$ $\Rightarrow\sqrt{6(3+ab+bc+ac)}=\sqrt{2.9+2.3(ab+bc+ac)}\leq2(a+b+c)$ So $N\leq2(a+b+c)$ $\Rightarrow P\geq\frac{a+b+c}{2}\geq\frac{3}{2}$ Using AM-GM inequality

P C - 5 years, 4 months ago

That's right

Department 8 - 5 years, 4 months ago

Olympiad for grade 10

The answer is 5.

1 solution

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