Do \(x;y;z>0\) và \(x^2+y^2+z^2=3\)

Nên \(0< x;y;z< \sqrt{3}\)

Ta có: \(\frac{1}{x+y+z}\le\frac{1}{9x}+\frac{1}{9y}+\frac{1}{9z}\)

\(\Rightarrow A\ge x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}-\frac{1}{9x}-\frac{1}{9y}-\frac{1}{9z}\)

\(\Leftrightarrow A\ge x+\frac{8}{9x}+y+\frac{8}{9y}+z+\frac{8}{9z}\)

Ta chứng minh: \(x+\frac{8}{9x}\ge\frac{x^2+33}{18}\)

\(\Leftrightarrow\left(x-1\right)^2\left(16-x\right)\ge\)

Do đó \(A\ge\frac{x^2+y^2+z^2+99}{18}=\frac{102}{18}=\frac{17}{3}\)

Dấu = xảy ra khi x=y=z=1

Dòng thứ 3 từ dưới lên là \(\left(x-1\right)^2\left(16-x\right)\ge0\)

                              Đúng do \(0< x< \sqrt{3}< 16\)

vì x+y+z=1nên

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

nen \(\frac{xy}{x^2+y^2}+\frac{yz}{y^2+z^2}+\frac{xz}{x^2+z^2}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) =\(\left(\frac{xy}{x^2+y^2}+\frac{x^2+y^2}{4xy}\right)+\left(\frac{yz}{y^2+z^2}+\frac{y^2+z^2}{4yz}\right)+\left(\frac{xz}{x^2+z^2}+\frac{x^2+z^2}{xz}\right)+\frac{3}{4}\)

\(\ge2.\frac{1}{2}+\frac{2.1}{2}+\frac{2.1}{2}+\frac{3}{4}=\frac{15}{4}\)(dpcm)

dau = xay ra khi x=y=z=1/3

Áp dụng BĐT Cauchy cho 3 số dương, ta được:

\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)

\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)

\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)

\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(đpcm\right)\)

Ta có: \(\frac{x+1}{y^2+1}=\left(x+1\right).\frac{1}{y^2+1}=\left(x+1\right)\left(1-\frac{y^2}{y^2+1}\right)\)

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

Thiết lập hai BĐT còn lại tương tự và cộng theo vế:

\(P\ge\left(x+y+z+3\right)-\frac{x\left(z+1\right)+y\left(x+1\right)+z\left(y+1\right)}{2}\)

\(=6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\) (*)

Lại có BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

Thật vậy,ta có: BĐT \(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)

Thay vào (*),ta có: \(P\ge6-\frac{\left(xy+yz+zx\right)+\left(x+y+z\right)}{2}\)

\(\ge6-\frac{\frac{\left(x+y+z\right)^2}{3}+3}{2}=6-\frac{3+3}{2}=3\)

Dấu "=" xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)

Vậy \(P_{min}=3\Leftrightarrow x=y=z=1\)

Bài t đúng 100% nhá,đứa nào tk sai t nhở? ngon vô làm lại=)

\(P=\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}\)

\(\ge3\sqrt[3]{\frac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\)

Mà theo BĐT AM - GM ta có tiếp:

\(xyz\le\left(\frac{x+y+z}{3}\right)^3=1\)

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\le\left(\frac{x+y+z+3}{3}\right)^3=8\)

\(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra tại x=y=z=1

Vậy..................

Cho a, b, c mà bắt chứng minh x, y, z nên ko chứng minh đc là đúng òi:)

\(VT-VP=\Sigma_{cyc}\frac{\left(x-y\right)^4}{4xy\left(x^2+y^2\right)}\ge0\)

a,b,c??? chỗ nào vậy bé ?? :)))

Cho các số thực dương a,b,c thỏa mãn x+y+z=1 .Chứng minh

Chỗ e in đậm.