bạn dùng BĐT Cauchuy-Swartch cho cs Bt thứ 2 là ra nhé

\(VT=\frac{1}{16}\left(\frac{1}{x}+\frac{4}{y}+\frac{16}{z}\right)\ge\frac{1}{16}\left(\frac{\left(1+2+4\right)^2}{x+y+z}\right)=\frac{49}{16}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{y}{2}=\frac{z}{4}\\x+y+z=1\end{matrix}\right.\) \(\Rightarrow\left(x;y;z\right)=\left(\frac{1}{7};\frac{2}{7};\frac{4}{7}\right)\)

Mấy cái dấu "=" anh tự xét.

Áp dụng BĐT AM-GM: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)

a) Áp dụng: \(VT\ge\frac{\left(a+b+c\right)^2}{3}.\frac{9}{2\left(a+b+c\right)}=\frac{3}{2}\left(a+b+c\right)\)

b) \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{x+y+z+3}=\frac{3}{4}\)

Ta đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c;\frac{1}{t}=d\)  ( a, b, c, d >0 )

Khi đó ta cần chứng minh:

 \(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)

\(VT=\frac{a^3}{\frac{b+c+d}{bcd}}+\frac{b^3}{\frac{a+c+d}{acd}}+\frac{c^3}{\frac{a+b+d}{abd}}+\frac{d^3}{\frac{a+b+c}{abc}}\)

\(=\frac{a^3}{\frac{a\left(b+c+d\right)}{abcd}}+\frac{b^3}{\frac{b\left(a+c+d\right)}{abcd}}+\frac{c^3}{\frac{c\left(a+b+d\right)}{abcd}}+\frac{d^3}{\frac{d\left(a+b+c\right)}{abcd}}\)

\(=\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{3\left(a+b+c+d\right)}=\frac{a+b+c+d}{3}=VP\)

Vậy ta đã chứng minh được

\(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)

Dấu "=" xảy ra <=> a = b = c = d 

Vậy : 

\(\frac{1}{x^3\left(yz+zt+ty\right)}+\frac{1}{y^3\left(xz+zt+tx\right)}+\frac{1}{z^3\left(xy+yt+tx\right)}+\frac{1}{t^3\left(xy+yz+zx\right)}\ge\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\)

Dấu "=" xảy ra <=> x = y = z = t = 1

\(\frac{x^4+y^4+z^4+t^4}{x^3+y^3+z^3+t^3}=\frac{\left(x^4+y^4+z^4+t^4\right)\left(x^2+y^2+z^2+t^2\right)}{\left(x^3+y^3+z^3+t^3\right)\left(x^2+y^2+z^2+t^2\right)}\)

\(\ge\frac{x^3+y^3+z^3+t^3}{x^2+y^2+z^2+t^2}=\frac{\left(x^3+y^3+z^3+t^3\right)\left(x+y+z+t\right)}{\left(x^2+y^2+z^2+t^2\right)\left(x+y+z+t\right)}\)

\(\ge\frac{x^2+y^2+z^2+t^2}{x+y+z+t}\ge\frac{\left(x+y+z+t\right)^2}{4\left(x+y+z+t\right)}=\frac{1}{4}\)

Dấu "=" xảy ra tại x=y=z=t=¹/4

Bài làm có tham khảo của GOD Đạt Hồ

Cách khác: (dù không biết đúng hay sai ạ!)

 BĐT \(\Leftrightarrow\frac{x^4+y^4+z^4+t^4}{x^3+y^3+z^3+t^3}\ge\frac{x+y+z+t}{4}\)

\(\Leftrightarrow4\left(x^4+y^4+z^4+t^4\right)\ge\left(x+y+z+t\right)\left(x^3+y^3+z^3+t^3\right)\)

Không mất tính tổng quát, giả sử \(x=min\left\{x,y,z\right\}\) và đặt \(x=a;y=a+u;z=a+v;t=a+w\) thì

\(a>0;u,v,w\ge0\). Ta có: 

\(VT-VP=\)

\(\ge0\).

Em là em nhìn sơ qua là em thấy BĐT cuối nó đúng đó ạ. Chị check lại thử

BĐT cần chứng minh tương đương với : \(\frac{\left(x+z\right)^2}{xz}\ge\frac{y\left(x+z\right)}{xz}+\frac{x+z}{y}\)

\(\Leftrightarrow\frac{x+z}{xz}\ge\frac{y}{xz}+\frac{1}{y}\Leftrightarrow y\left(x+z\right)\ge y^2+xz\)

\(\Leftrightarrow y^2-y\left(x+z\right)+xz\le0\Leftrightarrow\left(y-x\right)\left(y-z\right)\le0\) ( luôn đúng vì \(z\ge y\ge x>0\))

Vậy BĐT đã được chứng minh khi x = y = z

YLê Anh DuyPhùng Tuệ Minh Akai Haruma

đặt A=\(\frac{1}{x\left(x+1\right)}\) +\(\frac{1}{y\left(y+1\right)}\) +\(\frac{1}{z\left(z+1\right)}\)=\(\frac{1}{x}\)-\(\frac{1}{x+1}\)+\(\frac{1}{y}\)-\(\frac{1}{y+1}\)+\(\frac{1}{z}\)-\(\frac{1}{z+1}\)

Áp dụng BĐT phụ \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\) (bạn tự chứng minh nha,quy đồng ,nhân chéo ,chuyển về )⇒\(\frac{1}{a+b}\) ≤\(\frac{1}{4}\)(\(\frac{1}{a}\)+\(\frac{1}{b}\))

⇒A≥\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)-\(\frac{1}{4}\)(\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\)+3)

⇒A≥\(\frac{3}{4}\) (\(\frac{1}{x}\)+\(\frac{1}{y}\)+\(\frac{1}{z}\))-\(\frac{3}{4}\)≥\(\frac{3}{4}\) (\(\frac{9}{x+y+z}\))-\(\frac{3}{4}\)

⇒a≥\(\frac{9}{4}\)-\(\frac{3}{4}\)=\(\frac{3}{2}\) dpcm

dấu bằng xảy ra⇔x=y=z=1