\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)

\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\) 

\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)  

\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\) 

\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)

Dự đoán Min P=1 khi x+y+z=3

Đặt \(t=x+y+z\ge3\) 

\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\) 

\(\Rightarrow P\ge1\)

bạn là một thiên tài

\(Q=\Sigma\frac{x^4}{x^2+\sqrt{xy.zx}}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+xy+yz+zx}\ge\frac{x^2+y^2+z^2}{2}\ge\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x=y=z=1 

ta caàn chứng minh bđt 

\(\frac{x}{x+yz}+\frac{y}{y+zx}\ge\frac{x}{x+xz}+\frac{y}{y+yz}=\frac{1}{1+z}+\frac{1}{1+z}=\frac{2}{1+z}\)

tương tự + vào, dùng svác sơ

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 Min_P=\(\frac{15}{2}\)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 \(x=y=z=\frac{1}{2}\)

Một cách giải khác ( cách này em làm rùi giờ làm lại ạ ) cô Chi check em ạ :)

Áp dụng BĐT AM-GM ta có:

\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)

\(\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng tiếp BĐT AM-GM ta có:

\(y+\frac{1}{x}=y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\ge5\sqrt[5]{\frac{y}{256x^4}}\)

Tương tự \(z+\frac{1}{y}\ge5\sqrt[5]{\frac{z}{256y^4}};x+\frac{1}{z}\ge5\sqrt[5]{\frac{x}{256z^4}}\)

Sử dụng liên hoàn BĐT AM-GM ta có tiếp

\(P\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)

\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)

\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)

\(=15\sqrt[15]{\frac{1}{256^3\left(xyz\right)^3}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\left(\frac{x+y+z}{3}\right)^9}}\)

\(\ge15\sqrt[15]{256^3\cdot\frac{1}{2^9}}=\frac{15}{2}\)

Dấu "='" xảy ra tại x=y=z=¹/₂

Áp dụng BĐT AM-GM: $VP\leq \frac{25}{yz+zx+xy+4}$

Cần c/m: $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}$\leq \frac{25}{yz+zx+xy+4}$

$\Leftrightarrow (yz+zx+xy)(xy^{2}+yz^{2}+zx^{2})+4(xy^{2}+yz^{2}+zx^{2})\leq 25xyz+4(yz+zx+xy)+16$

BĐT trên sẽ được c/m nếu c/m được: $xy^{2}+yz^{2}+zx^{2}\leq 4$.

KMTTQ, g/sử y nằm giữa x và z. $\Rightarrow x(x-y)(y-z)\geq 0$

$\Leftrightarrow xy^{2}+yz^{2}+zx^{2}\leq y(x^{2}+xz+z^{2})\leq y(x+z)^{2}$

Đến đây áp dụng BĐT AM-GM:

$y(x+z)^{2}=4.y.(\frac{x+z}{2})(\frac{x+z}{2})\leq \frac{4(y+\frac{x+z}{2}+\frac{x+z}{2})^{3}}{27}=\frac{4(x+y+z)^{3}}{27}=4$ (đpcm)

Dấu bằng xảy ra khi, chẳng hạn $x=0;y=1;z=2$

Áp dụng BĐT AM-GM và BĐT Rearrangement  ta có:

\(VT=\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\)

\(=\frac{\left(x+y+z\right)^2+3\left(x+y+z\right)+xy^2+yz^2+zx^2+3}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)\(\le\frac{21+y\left(x+z\right)^2}{3\sqrt[3]{4\left(xy+yz+xz\right)}}\le\frac{21+\frac{\left(\frac{2\left(x+y+z\right)}{3}\right)^3}{2}}{3\sqrt[3]{4\left(xy+yz+zx\right)}}=\frac{21+4}{3\sqrt[3]{4\left(xy+yz+zx\right)}}=\frac{25}{3\sqrt[3]{4\left(xy+yz+zx\right)}}\)

Dấu "=" xảy ra <=> (x;y;z)=(2;1;0) và hoán vị của nó

xml hkjmf,gkjbhvn jbkvmcbnvdyjxnbv hjgfvchjwbfhyergfvyug h ghbf vchdsvhdc ghv eucbtrgvtcfrtfvgtcb tybk cjvh dgsx     hjutygfvhyfhefrd cr fb kosciugyrturikjht54tr273r6734vn cjhvdfbv dfjbgerutjh37347t567  t gn fvbrhkjbfghty 66u 67gfbrhtb vbnbdffrhg ';\ hvgn hvbhzxn cb gvfycbher 74y6t5rbfvnhsgt hbgvdhcvhjgey6t5u gewytdfjbxjhdv bn 6t5675t47t5648b   gryjhvdhybgfvdghv d vdfstrcdgvcc ghfvdshvh bbv3rt364tr  bgryjhvbnh vznhbbcv  nbmhfbvdghbv mhdfbdschmaewugugf ygvrfyug s g dg vyga4ut53746r87hyu  rf5ygygcsrbv sdbv x vc  bgyergty4gfytrfygtyfgrgyfyjugrfauygfugdv euygt674y4375y74