\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)

\(A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

\(A^2=4+2\sqrt{2^2-\sqrt{3}^2}\)

\(A^2=4+2\sqrt{4-3}=4+2=6\)

\(\Rightarrow A^2=6\Rightarrow A=\sqrt{6}\)

Select A

By AM-GM we have:

\(E=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)

\(=4\left(\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{4c}{a+b-c}\right)\)

\(\ge4\cdot3\sqrt[3]{\dfrac{abc}{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}\)

\(\ge4\cdot3\sqrt[3]{\dfrac{abc}{abc}}=4\cdot3=12\)

Need prove \(\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\le abc\)

The last inequality is right follow Schur's inequality

\("="\Leftrightarrow a=b=c\)

1.By Cauchy-Schwarz we have:

\(\left(1+1\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\)

Need prove \(\dfrac{\left(a^2+b^2\right)^2}{2}\ge a^2+b^2\Leftrightarrow\left(a^2+b^2\right)^2\ge2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+b^2\ge2\). By C-S again :

\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}=2\)

It's right or we are done :3

??? The word "l-i-k-e" is hidden ??? . The word " l-i-k-e" are many meanings why.Why hide it just because it means "thích" ( sorry i can't Find alternative words)?  It also has mean is "như là"

sorry :3. it's typo :3

- mustn't use  "Right mouse click" old OLM

uk about idea "Need to block the copy mode in this website." - i agree but it's not perhaps mustn't use  "Right mouse click" old OLM. It is very inconvenient

2. By Cauchy-Schwarz we have:

\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}\)

\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Similarly :\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{b+c}\right)\)

\(L.H.S\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}=R.H.S\)

This web too few questions, try to answer many question ? That's impossible !!

Yes but it's not mine, i need new method :)

Let \(x_1=\dfrac{a_2}{a_1};x_2=\dfrac{a_3}{a_2};...;x_n=\dfrac{a_1}{a_n}\)

We need prove \(\dfrac{a_1}{a_1+a_2+a_3}+\dfrac{a_2}{a_2+a_3+a_4}+...+\dfrac{a_n}{a_n+a_1+a_2}>1\)

It's right because \(n>3\) and \(a_i+a_{i+1}+a_{i+2}< a_1+a_2+...+a_n\forall i\)

CTV ? haha u need write it by English 

you need write source answer Câu hỏi của Kurosaki Akatsu - Toán lớp 9 - Học toán với OnlineMath

I think x=1; y=4; z=9 are solution. My idea comes from Py-ta-go theorem 

wrong tag : maxima-minima

From \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Hence \(A=\dfrac{a^3+b^3+c^3-3abc}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}=0\left(a+b+c=0\right)\)

Done  !!

Another way - Use Cauchy-Schwarz

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\ge\dfrac{9}{2}\)

\(\Leftrightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)

By Cauchy-Schwarz: \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{9}{2\left(a+b+c\right)}\)

\(\Rightarrow L.H.S=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)

\(\ge a+b+c\cdot\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}=R.H.S\)

The equality occurs for \(a=b=c\)

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+100}\)

\(=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{100\cdot\left(100+1\right)}{2}}\)

\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{100\cdot\left(100+1\right)}\)

\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{100\cdot\left(100+1\right)}\)

\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{100\cdot101}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{101}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)\(=2\left(\dfrac{1}{2}-\dfrac{1}{101}\right)=2\cdot\dfrac{99}{202}=\dfrac{99}{101}\)

@mathlove: \(3-x\ne x-3\) ???

Find minimize \(|m-2|+|m+3|\) ?

By inequality \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) we have:

\(\left|m-2\right|+\left|m+3\right|=\left|2-m\right|+\left|m+3\right|\)

\(\ge\left|2-m+m+3\right|=5\)

Done !

 If we let \(x = 1\) and \(o = – 1\), then note that consecutive symbols are replaced by their product. If we consider the product \(P \) of the nine values before and after each operation, we will see that the new \(P\) is the square of the old \(P\). Hence, \(P \) will always equal \(1 \) after an operation. So nine \(o\)'s yielding \(P = – 1\) can never happen.  

 In the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group.