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Ta có:
\(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\) (1)
Hiển nhiên suy ra được BĐT Am-Gm
Áp dụng (1) ta được:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};\frac{1}{y}+\frac{1}{z}\ge\frac{4}{y+z};\frac{1}{z}+\frac{1}{x}\ge\frac{4}{z+x}\)
Cộng các vế BĐT ta được
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\) (2)
Tương tự như vậy ta có:
\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{y+z}\ge2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\) (3)
Áp dụng (2) và (3) ta được:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge4\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Vậy Max A = 1
Đặt \(^{\hept{\begin{cases}x=a^2\\y=b^2\\z=c^2\end{cases}}\Rightarrow abc=1}\)
\(\Rightarrow P=\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\)
ÁP DỤNG BĐT AM-GM :
\(a^2+b^2\ge2ab\)
\(b^2+1\ge2b\)
\(\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2}.\frac{1}{ab+b+1}\)
Tương tự \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}.\frac{1}{bc+c+1}\)
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ac+a+1}\)
Cộng từng vế các bđt trên ta được
\(P\le\frac{1}{2}\)
Dấu "=" xảy ra khi x=y=z=1
Áp dụng bất đẳng thức Cauchy - Schwarz : \(\frac{a^2}{b}+\frac{c^2}{d}\ge\frac{\left(a+c\right)^2}{b+d}\)
\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1^2}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)(\(\left(a+b\right)^2\ge4a\))
Tương tự: \(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4};\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\)
\(\Rightarrow2.\left(\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\right)\ge4\left(\frac{x}{x^6+y^4}+\frac{y}{y^6+z^4}+\frac{z}{z^6+x^4}\right)\)
\(\Rightarrow\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)
với x,y,z >0 áp dụng bđt cosi ta có:
\(x^6+y^4>=2\sqrt{x^6y^4}=2x^3y^2\Rightarrow\frac{2x}{x^6+y^4}< =\frac{2x}{2x^3y^2}=\frac{1}{x^2y^2}\)
\(y^6+z^4>=2\sqrt{y^6z^4}=2y^3z^2\Rightarrow\frac{2y}{y^6+z^4}< =\frac{2y}{2y^3z^2}=\frac{1}{y^2z^2}\)
\(z^6+x^4>=2\sqrt{z^6x^4}=2z^3x^2\Rightarrow\frac{2z}{z^6+x^4}< =\frac{2z}{2z^3x^2}=\frac{1}{z^2x^2}\)
\(\Rightarrow\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}< =\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\left(1\right)\)
với x,y,z>0 áp dụng bđt cosi ta có:
\(\frac{1}{x^4}+\frac{1}{y^4}>=2\sqrt{\frac{1}{x^4}\cdot\frac{1}{y^4}}=\frac{2}{x^2y^2}\)
\(\frac{1}{y^4}+\frac{1}{z^4}>=2\sqrt{\frac{1}{y^4}\cdot\frac{1}{z^4}}=\frac{2}{y^2z^2}\)
\(\frac{1}{x^4}+\frac{1}{z^4}>=2\sqrt{\frac{1}{x^4}\cdot\frac{1}{z^4}}=\frac{2}{x^2z^2}\)
\(\Rightarrow\frac{2}{x^4}+\frac{2}{y^4}+\frac{2}{z^4}>=\frac{2}{x^2y^2}+\frac{2}{y^2z^2}+\frac{2}{x^2z^2}\Rightarrow\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}>=\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{x^2z^2}\)
\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{x^2z^2}< =\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\left(2\right)\)
từ \(\left(1\right)\left(2\right)\Rightarrow\frac{2x}{x^6+y^4}+\frac{2x}{y^6+z^4}+\frac{2x}{z^6+x^4}< =\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)(đpcm)
dấu = xảy ra khi x=y=z=1
Theo Cauche có:
\(\left(x+x+y+z\right)\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\sqrt[4]{x^2yz}.4\sqrt[4]{\frac{1}{x^2.y.z}}=16\)
=> \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\). Tương tự có:
\(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{x+2y+z}\) và \(\frac{2}{z}+\frac{1}{y}+\frac{1}{x}\ge\frac{16}{x+y+2z}\)
=> \(16.\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{2}{y}+\frac{1}{x}+\frac{1}{z}+\frac{2}{z}+\frac{1}{x}+\frac{1}{y}\)
\(16.\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le4.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=4.4=16\)
Chia cả 2 vế cho 16 => ĐPCM
Xét \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
<=> \(a^2+b^2\ge2ab\) (luôn đúng)
Dấu bằng xảy ra khi a=b
Áp dụng ta có
\(\frac{1}{x+3y}+\frac{1}{y+2z+x}\ge\frac{4}{2\left(x+2y+z\right)}=\frac{2}{x+2y+z}\)
\(\frac{1}{y+3z}+\frac{1}{z+2x+y}\ge\frac{2}{x+y+2z}\)
\(\frac{1}{z+3x}+\frac{1}{x+2y+z}\ge\frac{2}{2x+y+z}\)
Cộng các vế của các bđt trên
=> ĐPCM
Dấu bằng xảy ra khi x=y=z
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\)
\(=\frac{1}{\left(x+y\right)+\left(x+z\right)}+\frac{1}{\left(x+y\right)+\left(y+z\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\)
\(\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\("="\Leftrightarrow x=y=z=\frac{3}{4}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\)
\(\Rightarrow\frac{1}{16}.\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{2x+y+z}\)
CMTT: \(\frac{1}{x+2y+z}\le\frac{1}{16}.\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\), \(\frac{1}{x+y+2z}\le\frac{1}{16}.\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)
\(\Rightarrow\Sigma\frac{1}{2x+y+z}\le\frac{1}{16}.4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)=\frac{1}{16}.16=1\)
\(''=''\Leftrightarrow x=y=z=\frac{3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{5}{x+y+z}=\frac{x+y}{2z}=\frac{y+z-1}{2x}=\frac{z+x+1}{2y}=\frac{x+y+y+z-1+z+x+1}{2z+2x+2y}=1\)
=> x + y + z = 5 : 1 = 5 (1)
x + y = 2z (2)
y + z - 1 = 2x => y + z = 2x + 1(3)
z + x + 1 = 2y => x + z = 2y - 1(4)
Thay (2) vào (1) ta có:
2z + z =5
=> 3z = 5
=> z = 5 : 3 = 1,(6)
Thay (3) vào (1) ta có:
x + 2x + 1 = 5
=> 3x = 5 - 1 = 4
=> x = 4 : 3 = 1,(3)
=> 1,(3) + y + 1,(6) = 5
=> y + 3 = 5
=> y = 5 - 3 = 2
Vậy x = 1,(3) ; y = 2 ; z = 1,(6)
Mình là học sinh lớp 7 nên ko biết đúng ko
\(\frac{1}{3x+2y+z}=\frac{1}{x+x+x+y+y+z}\le\frac{1}{6^2}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Tương tự thì ta có:
\(\frac{1}{3x+2y+z}+\frac{1}{x+3y+2z}+\frac{1}{y+3z+2x}\)
\(\le\frac{1}{36}\left(\frac{3}{x}+\frac{2}{y}+\frac{1}{z}\right)+\frac{1}{36}\left(\frac{1}{x}+\frac{3}{y}+\frac{2}{z}\right)+\frac{1}{36}\left(\frac{1}{y}+\frac{3}{z}+\frac{2}{x}\right)\)
\(=\frac{6}{36}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{16}{6}=\frac{8}{3}\)
Dấu "=" xảy ra <=> x = y = z = 3/16
Ta có bất đẳng thức: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) với \(x,y>0\).
Dấu \(=\)xảy ra khi \(x=y\).
Ta có: \(\frac{1}{2x+y+z}=\frac{1}{x+y+x+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)
\(\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\).
Tương tự với hai số hạng còn lại.
Suy ra \(P\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)+\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)
\(=\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{2020}{4}=505\).
Dấu \(=\)xảy ra khi \(x=y=z=\frac{3}{2020}\).
Ta có:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};\frac{1}{y}+\frac{1}{x}\ge\frac{4}{x+y}\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\ge\frac{4}{x+y}+\frac{4}{x+y}\ge\frac{16}{x+2y+z}\Rightarrow\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)\(TT:\)
\(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right);\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)\\(S\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=1\)