We have: \(2^{2018}=\left(2^{1009}\right)^2=\left(2^{979}\cdot2^{30}\right)^2=\left(\left(2^{11}\right)^{89}\cdot2^{30}\right)^2\)

\(\equiv\left(48^{89}\cdot824\right)^2\left(mod1000\right)\)\(=\left(48^3\cdot\left(48^2\right)^{43}\cdot824\right)^2\)

\(\equiv\left(592\cdot304^{43}\cdot824\right)^2\left(mod1000\right)\)

\(\equiv\left(592\cdot\left(304^5\right)^8\cdot464\cdot824\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot24^8\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot776^2\right)^2\left(mod1000\right)\)

\(\equiv\left(912\cdot176\right)^2\left(mod1000\right)\)

\(\equiv512^2\left(mod1000\right)\equiv144\left(mod1000\right)\)

So the last three-digit numbers of \(M\) is \(144\)

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

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

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

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

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\sqrt{ab}\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\sqrt{ab}\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=\dfrac{\left(a-b\right)^2}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\)

Or \(Σ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+2\sqrt{ab}-b\right)}{\dfrac{1}{b}+\dfrac{1}{2\left(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}\right)}}\right)\ge0\)

Done !!

By AM-GM's ineq: \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

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

\(\LeftrightarrowΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}\left(\sqrt{ab}\right)\geΣ_{cyc}\sqrt{\dfrac{a^2+b^2}{2}}-Σ_{cyc}\left(\sqrt{ab}\right)\left(1\right)\)

We have: \(LHS_{\left(1\right)}\geΣ_{cyc}\dfrac{a^2}{b}-Σ_{cyc}a=Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\)

And \(RHS_{\left(1\right)}=Σ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

Or \(Σ_{cyc}\dfrac{\left(a-b\right)^2}{b}\geΣ_{cyc}\left(\dfrac{\left(a-b\right)^2}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{1}{b}-\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}\right)\right)\ge0\)

\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\dfrac{\dfrac{\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}+b\right)\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}-b\right)}{b^2\left(\sqrt{2\left(a^2+b^2\right)}+\sqrt{ab}\right)^2}}{\dfrac{1}{b}+\dfrac{1}{2\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{ab}}}\right)\ge0\)

Done !!!

\(Q=\dfrac{3\left(x+1\right)}{x^3+x^2+x+1}=\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{3}{x^2+1}\)

We have: \(x^2\ge0\forall x\Leftrightarrow x^2+1\ge1\forall x\)

\(\Leftrightarrow\dfrac{1}{x^2+1}\le1\forall x\)

\(\Leftrightarrow Q=\dfrac{3}{x^2+1}\le3\forall x\)

\("="\Leftrightarrow x=0\)

I WANT TO KNOW THE REASON WHY I CAN"T CHANGE MY AVATAR ? NEED A CLEARLY EXPLAIN AND WHO KNOW HOW TO CHANGE IT PLS SHARE WITH ME?

FORGIVE ME ABOUT SPAM BUT IT"S REALLY IMPORTANT WITH ME, Tks so much

\(\dfrac{x^2-z^2}{y+z}+\dfrac{y^2-x^2}{z+x}+\dfrac{z^2-y^2}{x+y}\ge0\)

\(\Leftrightarrow\dfrac{x^4+y^4+z^4-x^2y^2-y^2z^2-x^2z^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x^4-2x^2y^2+y^4\right)+\left(y^4-2y^2z^2+z^4\right)+\left(z^4-2x^2z^2+x^4\right)}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x^2-y^2\right)^2+\left(y^2-z^2\right)^2+\left(z^2-x^2\right)^2}{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge0\left(\text{*}\right)\)

The last inequality is obvious

\("="\Leftrightarrow x=y=z\)

By Cauchy-Schwarz's ineq:

\(VP=\dfrac{4a^2}{2ab+2ac}+\dfrac{4b^2}{2ab+2bc}+\dfrac{4c^2}{2ac+2bc}\)

\(\ge\dfrac{\left(2a+2b+2c\right)^2}{4\left(a+b+c\right)}=\dfrac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c=VT\)

When \(a=b=c=1 \Rightarrow A=3\)

\(a=b=1\rightarrow a+\frac{1}{b}>1\)

Hey con lợn "ghuy-pu" kia bớt đú đi nhé, mạng VN đã quá thừa trẻ trâu vào rồi, lag lắm rồi 

if right-handed-sign is \(a^2b^2c^2\), you can assume it.

Stop Touching YourSelf, please =((

( ͡° ͜ʖ ͡°) 

#NOTE:\(e=equation\)

\(e\left(2\right)\Leftrightarrow y=3-3x\)

\(\Rightarrow e\left(1\right)\Leftrightarrow32x^3-48x^2+30x+\left(4\left(3-3x\right)-7\right)\sqrt{1-\left(3-3x\right)}=7\)

\(\Leftrightarrow32x^3-48x^2+30x+\left(5-12x\right)\sqrt{3x-2}=7\)

\(\Leftrightarrow32x^3-48x^2+10x+6+\left(5-12x\right)\sqrt{3x-2}-\left(-20x+13\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(4x-3\right)\left(4x+1\right)+\dfrac{\left(x-1\right)\left(4x-3\right)\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)\left(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}\right)=0\)

See: \(2\left(4x+1\right)+\dfrac{\left(108x-73\right)}{\left(5-12x\right)\sqrt{3x-2}-20x+13}>0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=3-3x=3-3\cdot1=0\\y=3-3x=3-3\cdot\dfrac{3}{4}=\dfrac{3}{4}\end{matrix}\right.\)

KEITA FC 8C: TAO NGHI MAY KO NEN TU HOI TU TRA LOI :(

We have: \(2006=2005+1=x+1\)

Hence \(B=x^{2005}-2006x^{2004}+2006x^{2003}-...+2006x+1\)

\(=x^{2005}-\left(x+1\right)x^{2004}+\left(x+1\right)x^{2003}-...+\left(x+1\right)x+1\)

\(=x^{2005}-x^{2005}-x^{2004}+x^{2014}+x^{2013}-...+x^2+x+1\)

\(=x+1=2005+1=2006\)

By Cauchy-Schwarz's inequality we have:

\(VT=\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\)

\(=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(y+z\right)}\)

\(\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{x+y}\right)+\dfrac{1}{4}\left(\dfrac{z}{x+z}+\dfrac{z}{z+y}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}\right)=\dfrac{3}{4}\)

When \(x=y=z\)

ok you can post your solution which you think smaller than I =))

A compact answer ? You need to know that this SOS method is one of method COMPACT because it doesn't loss of variables. 

Let \(A=\sqrt{6+\sqrt{6+\sqrt{6+....}}}>0\)

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

\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)

\(\Rightarrow\left(A+2\right)\left(A-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}A-3=0\\A+2=0\end{matrix}\right.\)\(\Rightarrow A=3\left(A>0\right)\)

Can you make the next question more clearly ?