❤ѕѕѕσиɢσкυѕѕѕ❤

Ta có: \(x^2+y^2+z^2+t^2-\left(xy+yz+zt+tx\right)=1-1\)

\(\Leftrightarrow2\left(x^2+y^2+z^2+t^2-xy-yz-zt-tx\right)=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2+2t^2-2xy-2yz-2zt-tx=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zt+t^2\right)+\left(t^2-2tx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2=0\)

Vì \(\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0;\left(z-t\right)^2\ge0;\left(t-x\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-t\right)^2+\left(t-x\right)^2\ge0\)

Dấu "=" xảy ra khi x - y = 0 ; y - z = 0 ; z - t = 0 ; t - x = 0 <=> x = y = z = t

Khi đó \(x^2+y^2+z^2+t^2=x^2+x^2+x^2+x^2=4x^2=1\)

\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)

Vậy \(x=y=z=t=\pm\frac{1}{2}\)

Từ \(xyzt=1\) ta có: \(\dfrac{1}{x^3\left(yz+zt+ty\right)}=\dfrac{xyzt}{x^3\left(yz+zt+ty\right)}=\dfrac{yzt}{x^2\left(yz+zt+ty\right)}\)

Đánh giá tương tự ta có:

\(pt\Leftrightarrow\dfrac{yzt}{x^2\left(yz+zt+ty\right)}+\dfrac{xzt}{y^2\left(xz+zt+tx\right)}+\dfrac{xyt}{z^2\left(xy+yt+tx\right)}+\dfrac{xyz}{t^2\left(xy+yz+zx\right)}\ge3\left(yzt+xzt+xyt+xyz\right)=3yzt+3xzt+3xyt+3xyz\)

Ta sẽ chứng minh:

\(\dfrac{yzt}{x^2\left(yz+zt+ty\right)}\ge3yzt\). Cộng theo vế rồi suy ra đpcm

T gần đi học r,có gì tối về giải full cho

Áp dụng cauchy-schwarz:

\(VT=\sum\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)^2}{3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)}=VF\)

@Neet

\(VT=\dfrac{1}{x^3\left(yz+zt+ty\right)}+\dfrac{1}{y^3\left(xz+zt+tx\right)}+\dfrac{1}{z^3\left(xy+yt+tx\right)}+\dfrac{1}{t^3\left(xy+yz+xz\right)}\)

\(=\dfrac{\dfrac{1}{x^2}}{xyz+xzt+xyt}+\dfrac{\dfrac{1}{y^2}}{xyz+yzt+txy}+\dfrac{\dfrac{1}{z^2}}{xyz+yzt+ztx}+\dfrac{\dfrac{1}{t^2}}{xyt+yzt+txz}\)

\(=\dfrac{\dfrac{1}{x^2}}{\dfrac{xyz}{xyzt}+\dfrac{xzt}{xyzt}+\dfrac{xyt}{xyzt}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txy}{xyzt}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{ztx}{xyzt}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{xyt}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txz}{xyzt}}\)

\(=\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{t}+\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}}\)

\(\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)^2}{3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)}=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)=VP\)

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

đặt x/y=a hay xy/z=a hay j đó là ra nói chung là 4 biế
n lười nháp

Answer:

\(P=\frac{1}{1+x+xy+xyz}+\frac{1}{1+y+yz+yzt}+\frac{1}{1+z+zt+ztx}+\frac{1}{1+t+tx+txy}\)

\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+xyzt}+\frac{xy}{xy+xyz+xyzt+xyzt.x}+\frac{xyz}{xyz+xyzt+xyzt.x+xyzt.xy}\)

\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+1}+\frac{xy}{xy+xyz+1+x}+\frac{xyz}{xyz+1+x+xy}\)

\(=\frac{1+x+xy+xyz}{1+x+xy+xyz}\)

\(=1\)

Thay xyzt = 1 vào P, có:

P= \(\frac{x}{xyz+xy+x+xyzt\ }\) + \(\frac{y}{yzt+yz+y+1}+\frac{z}{xzt+zt+z+xyzt}+\frac{t}{xyt+tx+t+1}\)

\(P=\frac{x}{x.\left(yz+y+1+yzt\right)}+\frac{y}{yzt+yz+y+1}+\frac{z}{z.\left(xt+t+1+xyt\right)}+\frac{t}{xyt+tx+t+1}\)

\(P=\frac{1\ +y}{yz+y+yzt+1}\) \(+\frac{1+t}{xyt+tx+t+1}\)

\(P=\frac{1+y}{yz+y+yzt+xyzt\ }+\frac{1+t}{xyt+tx+t+1}\)

\(P=\frac{1+y}{y.z.\left(xyt+tx+t+1\right)}+\frac{yz+tyz}{yz.\left(xyt+tx+t+1\right)}\)

\(P=\frac{1+y+yz+tyz}{yz.\left(xyt+tx+t+1\right)}=\frac{1+y+yz+tyz}{xyzt.\left(1+y+yz+tyz\right)}=\frac{1}{xyzt}=1\)

KL: P = 1 tại xyzt=1

\(\dfrac{x}{xyz+xy+x+1}+\dfrac{y}{yzt+yz+y+1}+\dfrac{z}{xzt+zt+z+1}+\dfrac{t}{xyt+tx+t+1}\)

= \(\dfrac{x}{xyz+xy+x+1}+\dfrac{xy}{xyzt+xyz+xy+x}+\dfrac{xyz}{x^2yzt+xyzt+xyz+xy}+\dfrac{xyzt}{x^{2^{ }}y^2zt+x^2yzt+xyzt+xyz}\)

= \(\dfrac{x}{xyz+xy+x+1}+\dfrac{xy}{1+xyz+xy+x}+\dfrac{xyz}{x+1+xyz+xy}+\dfrac{1}{xy+x+1+xyz}\)

= \(\dfrac{x+xy+xyz+1}{x+xy+xyz+1}\)

= 1

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\\\dfrac{y}{x}+1+\dfrac{y}{z}=0\\\dfrac{z}{x}+\dfrac{z}{y}+1=0\end{matrix}\right.\\ \Rightarrow\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=-3\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\dfrac{yz+xz+xy}{xyz}=0\\ \Rightarrow yz+xz+xy=0\)

\(\Rightarrow\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\left(xy+xz+yz\right)=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)

\(\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(\dfrac{-1}{z}\right)^3\)

\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3\dfrac{1}{x}\dfrac{1}{y}\dfrac{-1}{z}\)

\(\Leftrightarrow\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)xyz=3\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}.xyz\)

\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)

Vì 1/x + 1/y + 1/z = 0 nên lần lượt nhân vs x; y; z ta có:
1 + x/y + x/z = 0 (1)
1 + y/z + y/x = 0 (2)
1 + z/x + z/y = 0 (3)
Từ (1); (2); (3) suy ra : x/y + y/z + z/x + x/z + y/x + z/y = - 3 (*)
Mặt khác : 1/x + 1/y + 1/z = 0 nên quy đồng lên ta có:
(xy + yz + zx)/xyz = 0 hay xy + yz + zx = 0
Hay : (1/x^2 + 1/y^2 + 1/z^2).(xy + yz + zx) = 0
khai triển ra :
yz/x^2 + zx/y^2 + xy/z^2 + x/y + y/z + z/x + x/z + y/x + z/y = 0
Vậy : yz/x^2 + zx/y^2 + xy/z^2 = - (x/y + y/z + z/x + x/z + y/x + z/y) = 3 (theo (*))