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Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
Gọi cái biểu thức đó là P nha
Trước tiên chứng minh:
\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\left(\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right)=0\)
\(\Leftrightarrow\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\Leftrightarrow x-y+y-z+z-x=0\)( đúng )
Giờ ta quay lại bài toán ban đầu
Ta có:
\(\Leftrightarrow2P=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)
\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)
\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}=\frac{1}{2}\)
\(\Rightarrow P\ge\frac{1}{4}\)
bài này cần x,y,z>0 nữa, vừa xem xong bài y hệt của LCC :v
Dự đoán dấu "=" khi \(x=y=z=1\) thì \(P=24\)
Ta chứng minh P=24 là GTNN
Thật vậy áp dụng BĐT C-S ta có:
\(P=Σ\frac{\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2}{\left(z^2+1\right)\left(x+y\right)^2}\ge\frac{\left(Σ\left(x+1\right)\left(y+1\right)\left(x+y\right)\right)^2}{Σ\left(z^2+1\right)\left(x+y\right)^2}\)
Cần chứng minh: \(\frac{\left(Σ\left(x+1\right)\left(y+1\right)\left(x+y\right)\right)^2}{Σ\left(z^2+1\right)\left(x+y\right)^2}\ge24\)
\(\Leftrightarrow\left(Σ\left(x+1\right)\left(y+1\right)\left(x+y\right)\right)^2\ge24Σ\left(z^2+1\right)\left(x+y\right)^2\)
Đặt \(\hept{\begin{cases}x+y+z=3u\\xy+yz+xz=3v^2\\xyz=w^3\end{cases}}\) \(\Rightarrow u=1\) thì
\(Σ\left(x+1\right)\left(y+1\right)\left(z+1\right)=Σ\left(x^2y+x^2z+2x^2+2xy+2x\right)\)
\(=9uv^2-3w^3+2u\left(9u^2-6v^2\right)+9uv^2+6u^3=3\left(8u^3+uv^2-w^3\right)\)
Và \(Σ\left(z^2+1\right)\left(x+y\right)^2=2Σ\left(x^2y^2+x^2yz+x^2u+xyu^2\right)\)
\(=2\left(9v^4-6uw^3+3uw^3+9u^4-6u^2v^2+3u^2v^2\right)\)
\(=6\left(3u^4-u^2v^2+3v^4-uw^3\right)\). Can cm \(f\left(w^3\right)\ge0\)
\(f\left(w^3\right)=\left(8u^3+uv^2-w^3\right)^2-16\left(3u^6-u^4v^2+3u^2v^4-u^3w^3\right)\)
\(f'\left(w^3\right)=-2\left(8u^3+uv^2-w^3\right)+16u^3=2w^3-2uv^2\le0\)
Thay \(f\) la ham` ngh!ch bien, do đó, BĐT có 1 GTLN của w3 khi 2 biến bằng nhau
Đặt \(y=x;z=3-2x\), Khi đó:
\(BDT\Leftrightarrow\left(x-1\right)^2\left(x^4-2x^3-11x^2+24x+4\right)\ge0\)
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)
=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)
=>\(P>=2\sqrt{2}\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)
Dấu = xảy ra khi x=y=z=6căn 2
dễ mà bạn :))) gáy tí , sai thì thôi
\(P=\frac{x^3}{\left(1+x\right)\left(1+y\right)}+\frac{y^3}{\left(1+y\right)\left(1+z\right)}+\frac{z^3}{\left(1+z\right)\left(1+x\right)}\)
\(=\frac{x^3\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}+\frac{y^3\left(1+x\right)}{\left(1+y\right)\left(1+x\right)\left(1+z\right)}+\frac{z^3\left(1+y\right)}{\left(1+x\right)\left(1+z\right)\left(1+y\right)}\)
\(=\frac{x^3\left(1+z\right)+y^3\left(1+x\right)+z^3\left(1+y\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\frac{3\sqrt[3]{x^3y^3z^3\left(1+x\right)\left(1+y\right)\left(1+z\right)}}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)
đến đây áp dụng BĐT phụ ( 1+a ) ( 1+b ) ( 1+c ) >= 8abc
EZ :)))
\(Q\ge2\left(x+y+z\right)+3.\frac{9}{x+y+z}=2\left(x+y+z\right)+\frac{27}{x+y+z}.\)
Đặt X+Y+Z=t (\(t\le1\))
\(Q\ge2t+\frac{27}{t}=\left(2t+\frac{2}{t}\right)+\frac{25}{t}\ge2\sqrt{2t.\frac{2}{t}}+\frac{25}{1}=4+25=29\\ \)
Dấu = xảy ra khi x=y=z=1/3
Theo bđt cô si ta có : \(x+y+z\ge3\sqrt[3]{xyz}\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
=> \(Q\ge6\sqrt[3]{xyz}+9\sqrt[3]{\frac{1}{xyz}}\ge2\sqrt{6\sqrt[3]{xyz}\cdot9\sqrt[3]{\frac{1}{xyz}}}=6\sqrt{6}\)
Dấu = xảy ra khi : \(6\sqrt[3]{xyz}=9\sqrt[3]{\frac{1}{xyz}}\) Giải ra ta đc : \(xyz=\frac{3}{2}\sqrt{\frac{3}{2}}\)
Ta có:
\(H=\frac{1}{x^3\left(y+z\right)}+\frac{1}{y^3\left(z+x\right)}+\frac{1}{z^3\left(x+y\right)}\)
\(=\frac{\frac{1}{x^2}}{x\left(y+z\right)}+\frac{\frac{1}{y^2}}{y\left(z+x\right)}+\frac{\frac{1}{z^2}}{z\left(x+y\right)}\)
\(=\frac{\left(\frac{1}{x}\right)^2}{xy+zx}+\frac{\left(\frac{1}{y}\right)^2}{yz+xy}+\frac{\left(\frac{1}{z}\right)^2}{zx+yz}\)
Áp dụng BĐT Bunyakovsky dạng cộng mẫu ta được:
\(H\ge\frac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{2\left(xy+yz+zx\right)}=\frac{\left(\frac{xy+yz+zx}{xyz}\right)^2}{2\left(xy+yz+zx\right)}=\frac{\left(xy+yz+zx\right)^2}{2\left(xy+yz+zx\right)}\)
\(=\frac{xy+yz+zx}{2}\ge\frac{3\sqrt[3]{\left(xyz\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: x = y = z = 1
Vậy Min(H) = 3/2 khi x = y = z = 1
Đặt \(\left(x;y;z\right)=\left(a;b;c\right)\) (em ko có ý gì cả, chỉ là gõ quen tay hơn thôi:V)
Đặt \(p=a+b+c;q=ab+bc+ca;r=abc\)
Quy vể: Tìm min, max của P = p biết p, q, r > 0 thỏa mãn \(p^2-2q-p\le\frac{4}{3}\)
Ta có: \(\frac{4}{3}\ge p^2-2q-p\ge\frac{1}{3}p^2-p\)
Do đó \(\frac{1}{3}p^2-p-\frac{4}{3}\le0\Leftrightarrow-1\le p\le4\)
Do đó....
P/s: đúng ko ạ?