\(4\left(x^2+4x+2\right)=\) \(11\sqrt{x^4+4}\)

\(4^2\left(x^2+4x+2\right)^2\)\(=\)\(11^2.\left(x^4+4\right)\)

\(16.\)\(\left(x^4+16x^2+4+8x^3+4x^2+16x\right)\)\(=121x^4+484\)

\(16.\left(x^4+8x^3+20x^2+16x+4\right)\)\(=\)\(121x^4+484\)

\(16x^4+128x^3+320x^2+256x+64\)\(=\)\(121x^4+484\)

\(121x^4+484\)\(=\)\(16x^4+128x^3+320x^2+256x+64\)

\(105x^4+420\)\(=\)\(128x^3+320x^2+256x\)

\(105x^4-128x^3-320x^2-256x+420\)\(=0\)

\(4\left(x^2+4x+2\right)=11\sqrt{x^4+4}\)

\(\Leftrightarrow\left[4\left(x^2+4x+2\right)\right]^2=\left(11\sqrt{x^4+4}\right)^2\)

\(\Leftrightarrow-105x^4+128x^3+320x^2+256x-420=0\)

\(\Leftrightarrow\left(-3x^2+10x-6\right)\left(35x^2+74x+40\right)=0\)

\(\Leftrightarrow x=\frac{5\pm\sqrt{7}}{3}\) 

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\)

Áp dụng BĐT Cauchy-schwars ta có: 

\(\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2.\left(ab+bc+ca\right)}\)

Ta chứng minh BĐT phụ:

 \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)

\(\Leftrightarrow3.\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)

\(\Leftrightarrow3ab+3bc+3ca\le a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( BĐT luôn đúng )

Vậy \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)

Áp dụng: 

\(\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2.\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}.\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)

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

                                                      đpcm

Tham khảo nhé~

shinichi ơi giải giùm mk bài toán vs

\(\sqrt{49}=7\)

Vì 7² = 49

Sửa : \(\pm\sqrt{49}=\pm7\)

Lúc nãy làm nhầm 

với a,b,x,y không âm ta có

a,\(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)

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

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

Xet \(xy-2\ge0\) thì co hệ

\(\hept{\begin{cases}xy-2=4-y^2\\x^2-xy+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y^2+xy=6\\6x^2-6xy=-6\end{cases}}\)

Lây trên trừ dươi được

\(y^2-5xy+6x^2=0\)

\(\Leftrightarrow\left(y-3x\right)\left(y-2x\right)=0\)

Xet Xet \(xy-2< 0\) thì co hệ

\(\hept{\begin{cases}2-xy=4-y^2\\x^2-xy+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y^2-xy=2\\x^2-xy=-1\end{cases}}\)

Lây trên cộng đươi được

\(\left(x-y\right)^2=1\)

Làm nôt