\(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+zx}+\dfrac{z}{z^2+xy}\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}+\dfrac{1}{\sqrt{xy}}\right)\le\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

? cho a,b,c tìm x,y,z là seo?

chắc đề cho x+y+z=1

\(=>\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

\(=>\dfrac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}\)

\(=\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

làm tương tự với \(\dfrac{y}{y+\sqrt{y+xz}},\dfrac{z}{z+\sqrt{z+xy}}\)

\(=>A\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) dấu"=" xảy ra<=>x=y=z=`/3

Ta có:

\(VT=\dfrac{x^2}{x^3-xyz-2013x}+\dfrac{y^2}{y^3-xyz-2013y}+\dfrac{z^2}{z^3-xyz-2013z}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013.\left(z+y+z\right)}\)

\(VT=\dfrac{\left(x+y+x\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right).\left(xy+yz+xz\right)-xyz\right]}\)

\(VT=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}\)

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

\(\Rightarrow\) Đpcm.

Với a; b dương, nếu \(a\ge b\) thì \(\dfrac{1}{a}\le\dfrac{1}{b}\)

Áp dụng BĐT Cô-si cho mẫu số vế trái ta được:

\(\dfrac{1}{x^2+yz}+\dfrac{1}{y^2+xz}+\dfrac{1}{z^2+xy}\le\dfrac{1}{2x\sqrt{yz}}+\dfrac{1}{2y\sqrt{xz}}+\dfrac{1}{2z\sqrt{xy}}\)

\(\Rightarrow VT\le\dfrac{\sqrt{yz}}{2xyz}+\dfrac{\sqrt{xz}}{2xyz}+\dfrac{\sqrt{xy}}{2xyz}=\dfrac{\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}{2xyz}\)

Tiếp tục dùng Cô-si cho tử số:

\(VT\le\dfrac{\dfrac{y+z}{2}+\dfrac{x+z}{2}+\dfrac{x+y}{2}}{2xyz}=\dfrac{x+y+z}{2xyz}\)

\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\) (đpcm)

Dấu "=" xảy ra khi x=y=z

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

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

Gọi cái thiệt gớm đó là P

Ta có:

\(xy+yz+zx=xyz\)

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

Ta có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64y}+\dfrac{1+y}{64x}\ge3\sqrt[3]{\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}.\dfrac{1+x}{64y}.\dfrac{1+y}{64x}}=\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{64x}-\dfrac{1}{64y}-\dfrac{1}{32}\left(1\right)\)

Tương tự ta cũng có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{64y}-\dfrac{1}{64z}-\dfrac{1}{32}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{64z}-\dfrac{1}{64x}-\dfrac{1}{32}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta được

\(P\ge\dfrac{3}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Đặt cái ban đầu là P

Ta có: \(xy+yz+zx=xyz\)

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

Ta lại có:

\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)

\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)

Tương tự ta có:

\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)

Từ (1), (2), (3) ta có:

\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)

\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)

Dấu = xảy ra khi \(x=y=z=3\)

Lời giải:

Ta xét hiệu sau:

\(x^3+y^3-xy(x+y)=x^3-x^2y-(xy^2-y^3)\)

\(=x^2(x-y)-y^2(x-y)=(x^2-y^2)(x-y)=(x-y)^2(x+y)\geq 0, \forall x,y>0\)

\(\Rightarrow x^3+y^3\geq xy(x+y)(*)\)

\(\Rightarrow x^3+y^3+xy\geq xy(x+y+1)\)

\(\Rightarrow \frac{xy}{x^3+y^3+xy}\leq \frac{xy}{xy(x+y+1)}=\frac{1}{x+y+1}\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế, suy ra:

\(\text{VT}\leq \underbrace{\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{x+z+1}}_{M}(1)\)

Vì $xyz=1$ nên tồn tại $a,b,c>0$ sao cho \((x,y,z)=(\frac{a^2}{bc}, \frac{b^2}{ac}, \frac{c^2}{ab})\)

Khi đó:

\(M=\frac{abc}{a^3+b^3+abc}+\frac{abc}{b^3+c^3+abc}+\frac{abc}{c^3+a^3+abc}\)

\(\leq \frac{abc}{ab(a+b)+abc}+\frac{abc}{bc(b+c)+abc}+\frac{abc}{ca(c+a)+abc}\) (áp dụng công thức $(*)$)

hay \(M\leq \frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=\frac{a+b+c}{a+b+c}=1(2)\)

Từ \((1);(2)\Rightarrow \text{VT}\leq 1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$

Bài của chị Akai đoạn đầu hơi phức tạp(em nghĩ thế).

Ta có:

\(\left(x-y\right)^2\ge0\) với \(\forall x,y\)

\(\Rightarrow x^2+y^2-xy\ge0\) với \(\forall x,y\)

\(\Rightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)với\(\forall x,y\)

\(\Rightarrow x^3+y^3\ge xy\left(x+y\right)\) với \(\forall x,y\)

Rồi giải tiếp như chị ấy.