Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)

\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)

Ta có \(abc\ge1\)

\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)

\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)

Lời giải:

Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)

BĐT cần chứng minh trở thành:

\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)

Thật vậy, áp dụng BĐT Cauchy ta có:

\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)

\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)

\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)

Cộng theo vế các BĐT trên và rút gọn :

\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)

\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)

Vậy \((*)\) được chứng minh. Bài toán hoàn tất.

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)

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 \(x=y=z=3\)

\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)

\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)

\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)

\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)

\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)

\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)

\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)

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

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

a) 3.(ab+bc+ac)≤a²+b²+c²+2ab+2bc+2ac

<=> \(a^2+b^2+c^2-ab-bc-ac\ge0\)

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

<=> (a-b)²+(b-c)²+(a-c)²≥0 ( luon dung voi moi a,b,c)

b) ap dung ket qua tren va vế sau bn xem bài giải của mk ở trên

Ta có nhận xét sau:

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

Tương tự với các phân thức còn lại

Ta đặt:

     \(\left\{{}\begin{matrix}a=xy\\b=yz\\c=zx\end{matrix}\right.\)

     \(\Rightarrow abc=1\) và \(a,b,c>0\)

Biểu thức P trở thành:

     \(P=\Sigma_{cyc}\dfrac{a}{b+c}+2\Sigma_{cyc}\dfrac{a^2}{b+c}\)

Dễ thấy:

     \(\Sigma_{cyc}\dfrac{a}{b+c}\ge\dfrac{3}{2}\) (Nesbit)

     \(\Sigma_{cyc}\dfrac{a^2}{b+c}\ge\dfrac{a+b+c}{2}\ge\dfrac{3\sqrt[3]{abc}}{2}=\dfrac{3}{2}\)

Do đó:

     \(P\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Bài 1:

\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )

Khi đó:

\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)

\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)

\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)

(Áp dụng BĐT Cauchy_Schwarz)

Theo BĐT Cauchy dễ thấy:

\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)

\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$

Bài 2:

Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)

Ta có:

\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)

\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)

Áp dụng BĐT Cauchy:

\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)

\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)

\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)

Nhân theo vế:

\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)

\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)

\(\Rightarrow \text{VT}\leq 1\) (đpcm)

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

\(\)