Ta có: \(\frac{a}{\left(a+1\right)\left(b+1\right)}+\frac{b}{\left(b+1\right)\left(c+1\right)}+\frac{c}{\left(c+1\right)\left(a+1\right)}\ge\frac{3}{4}\)\(\Leftrightarrow\frac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\frac{3}{4}\)\(\Leftrightarrow4\left(ab+bc+ca\right)+4\left(a+b+c\right)\ge3abc+3\left(ab+bc+ca\right)+3\left(a+b+c\right)+3\)\(\Leftrightarrow ab+bc+ca+a+b+c\ge6\)(abc = 1)

Bất đẳng thức cuối đúng theo bất đẳng thức Cô - si nên ta có điều phải chứng minh

Đẳng thức xảy ra khi a = b = c = 1

Ta có: \(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=2\Leftrightarrow\left(1-\frac{a}{a+1}\right)+\left(1-\frac{b}{b+1}\right)+\left(1-\frac{c}{c+1}\right)=1\)\(\Leftrightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1\)\(\Leftrightarrow\left(b+1\right)\left(c+1\right)+\left(c+1\right)\left(a+1\right)+\left(a+1\right)\left(b+1\right)=\left(a+1\right)\left(b+1\right)\left(c+1\right)\)\(\Leftrightarrow a+b+c+2=abc\ge3\sqrt[3]{abc}+2\)(Bất đẳng thức Cô - si)

Đặt \(t=\sqrt[3]{abc}\)thì \(t^3\ge3t^3+2\Leftrightarrow\left(t-2\right)\left(t+1\right)^2\ge0\Leftrightarrow t\ge2\)(Do \(\left(t+1\right)^2>0\forall t>0\))

\(\Rightarrow abc\ge8\)

\(\Rightarrow ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\ge12\)(đpcm)

Đẳng thức xảy ra khi a = b = c = 2

Ta có:

\(\left(a+1\right)^2\left(b+1\right)^2=\left[\left(a+1\right)\left(b+1\right)\right]^2=\left(1+a+b+ab\right)^2\)

\(=\left[\left(ab+1\right)+\left(a+b\right)\right]^2\ge4\left(a+b\right)\left(ab+1\right)\)

\(=4a^2b+4ab^2+4a+4b=\left(4a^2b+4b\right)+\left(4ab^2+4a\right)\)

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

\(\Rightarrow\frac{\left(a+1\right)^2\left(b+1\right)^2}{1+c^2}\ge\frac{4a\left(1+b^2\right)}{1+c^2}+\frac{4b\left(1+a^2\right)}{1+c^2}\)

Tương tự ta chứng minh được:
\(\frac{\left(b+1\right)^2\left(c+1\right)^2}{1+a^2}\ge\frac{4c\left(1+b^2\right)}{1+a^2}+\frac{4b\left(1+c^2\right)}{1+a^2}\)

\(\frac{\left(a+1\right)^2\left(c+1\right)^2}{1+b^2}\ge\frac{4a\left(1+c^2\right)}{1+b^2}+\frac{4c\left(1+a^2\right)}{1+b^2}\)

Cộng vế 3 BĐT trên lại ta được:

\(VT\ge4a\left(\frac{1+b^2}{1+c^2}+\frac{1+c^2}{1+b^2}\right)+4b\left(\frac{1+a^2}{1+c^2}+\frac{1+c^2}{1+a^2}\right)+4c\left(\frac{1+a^2}{1+b^2}+\frac{1+b^2}{1+a^2}\right)\)

\(\ge8a+8b+8c=8\left(a+b+c\right)=8\cdot3=24\) (BĐT Cauchy)

Dấu "=" xảy ra khi: a = b = c = 1

Áp dụng bất đẳng thức AM - GM, ta được: 

\(\left(a+1\right)^2\left(b+1\right)^2=\left(ab+1+a+b\right)^2\ge4\left(ab+1\right)\left(a+b\right)=4a\left(1+b^2\right)+4b\left(1+a^2\right)\)\(\Rightarrow\frac{\left(a+1\right)^2\left(b+1\right)^2}{1+c^2}\ge4a.\frac{1+b^2}{1+c^2}+4b.\frac{1+a^2}{1+c^2}\)

Câu nèo thé ?_?

Câu nào thế bạn????

Ta có: \(\hept{\begin{cases}x^2-xy+y^2=8\\x^2+3xy+y^2=15\end{cases}}\Leftrightarrow\hept{\begin{cases}4xy=7\\x^2-xy+y^2=8\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{7}{4y}\\x^2-xy+y^2=8\end{cases}}\) thay vào ta được:

\(\left(\frac{7}{4y}\right)^2-\frac{7}{4}+y^2=8\Leftrightarrow\frac{49}{16y^2}+y^2=\frac{39}{4}\)

\(\Leftrightarrow\frac{16y^4+49}{16y^2}=\frac{39}{4}\Leftrightarrow16y^4+49=156y^2\)

\(\Leftrightarrow16y^4-156y^2+49=0\)

\(\Leftrightarrow\orbr{\begin{cases}y^2=\frac{39+5\sqrt{53}}{8}\\y^2=\frac{39-5\sqrt{53}}{8}\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=\sqrt{\frac{39+5\sqrt{53}}{8}}\Rightarrow x=\frac{7}{4\sqrt{\frac{39+5\sqrt{53}}{8}}}\\y=\sqrt{\frac{39-5\sqrt{53}}{8}}\Rightarrow x=\frac{7}{4\sqrt{\frac{39-5\sqrt{53}}{8}}}\end{cases}}\)

Vậy HPT có 2 nghiệm (x;y) thỏa mãn:

\(\left(\frac{7}{4\sqrt{\frac{39+5\sqrt{53}}{8}}};\sqrt{\frac{39+5\sqrt{53}}{8}}\right);\left(\frac{7}{4\sqrt{\frac{39-5\sqrt{53}}{8}}};\sqrt{\frac{39-5\sqrt{53}}{8}}\right)\)

Em cũng bị sau đó thì tìm zalo cô giáo để gửi bài thui nè