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1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm
Sau vài phút cố gắng thì khẳng định đề bài của em bị sai
\(x^2+2y^2+3=x^2+y^2+y^2+1+2\ge2xy+2y+2\)
\(z^2+2x^2+3\ge2zx+2x+2\)
\(y^2+2z^2+3\ge2yz+2z+2\)
Dễ chứng minh được \(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}=1\)
\(\Rightarrow\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{z^2+2x^2+3}+\dfrac{1}{y^2+2z^2+3}\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Bài 1
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
\(M=\dfrac{x+12-15}{x}+\dfrac{y+12-15}{y}+\dfrac{z+12-15}{z}\)
\(M=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-3}{z}\)
\(M=1-\dfrac{3}{x}+1-\dfrac{3}{y}+1-\dfrac{3}{z}\)
\(M=3-\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)\)
\(M=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}=\dfrac{3}{4}\)
\(\Rightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{9}{4}\)
\(\Rightarrow3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow M\le\dfrac{3}{4}\)
Vậy \(M_{max}=\dfrac{3}{4}\)
Dấu " = " xảy ra khi \(x=y=z=4\)
Bài 2
\(P=\dfrac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}+\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\)
Xét \(\dfrac{a^3+b^3+c^3}{4abc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{4abc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}=\dfrac{9}{ab+bc+ca}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2-ab-bc-ca\right)}{4\left(ab+bc+ca\right)}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)-9\left(ab+bc+ca\right)}{4\left(ab+bc+ca\right)}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{9}{4}+\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{bc}+\dfrac{1}{ca}+\dfrac{1}{ab}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+\dfrac{3}{4}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{4abc}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\ge\dfrac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^3+b^3+c^3}{4abc}-\dfrac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}-\dfrac{3}{2}\) (1)
Xét \(\dfrac{\left(a+b+c\right)^2}{30\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)}{30\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{1}{30}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\) (2)
Cộng (1) và (2) theo từng vế
\(P\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}-\dfrac{22}{15}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge2\sqrt{\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{225\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}}\)
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge2\sqrt{\dfrac{1}{225}}\)
\(\Rightarrow\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\ge\dfrac{2}{15}\)
\(P\ge\dfrac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}+\dfrac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}-\dfrac{22}{15}\ge\dfrac{2}{15}-\dfrac{22}{15}=-\dfrac{4}{3}\)
\(\Leftrightarrow P\ge-\dfrac{4}{3}\)
Vậy \(P_{min}=\dfrac{-4}{3}\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
Điều kiện: \(\left\{ \begin{array}{l} x > - 2\\ y > 1\\ x + y > 0 \end{array} \right.\)
Hệ phương trình tương đương: \(\left\{ \begin{array}{l} \sqrt {\dfrac{{x + y}}{{x + 2}}} + \sqrt {\dfrac{{x + y}}{{y - 1}}} = 2\\ {\left( {\dfrac{{x + 2}}{{x + y}}} \right)^2} + \left( {\dfrac{{y - 1}}{{x + y}}} \right)^2 = 2 \end{array} \right.\). Đặt \(\left\{ \begin{array}{l} a = \sqrt {\dfrac{{x + y}}{{x + 2}}} \\ b = \sqrt {\dfrac{{x + y}}{{y - 1}}} \end{array} \right.\) (với \(a,b > 0\))
Ta có hệ phương trình: \(\left\{ \begin{array}{l} a + b = 2\\ \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} = 2 \end{array} \right.\left( * \right)\)
Áp dụng BĐT AM - GM, ta có:
\(\begin{array}{l} 2 = a + b \geqslant 2\sqrt {ab} \Rightarrow ab \leqslant 1\\ 2 = \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} \geqslant 2\sqrt {\dfrac{1}{{{a^4}}}.\dfrac{1}{{{b^4}}}} \Rightarrow ab \geqslant 1 \end{array}\)
Thế nên \(\left( * \right) \Leftrightarrow a = b = 1\)
Ta lại có hệ phương trình: \(\left\{ \begin{array}{l} \dfrac{{x + y}}{{x + 2}} = 1\\ \dfrac{{x + y}}{{y - 1}} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - 1\\ y = 2 \end{array} \right.\)
Vậy hệ phương trình có nghiệm là \((-1;2)\)
Đk: \(\left\{{}\begin{matrix}x>-2\\y>1\\x+y>0\end{matrix}\right.\)
hpt\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\2\left(x+y\right)^2=\left(x+2\right)^2+\left(y-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\\left(\dfrac{x+2}{x+y}\right)^2+\left(\dfrac{y-1}{x+y}\right)^2=2\end{matrix}\right.\)
Đặt \(a=\sqrt{\dfrac{x+y}{x+2}},b=\sqrt{\dfrac{x+y}{y-1}}\left(a,b>0\right)\)
Ta có hệ: \(\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4+b^4=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=2a^4b^4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(4-2ab\right)^2-2a^2b^2=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4b^4=a^2b^2-8ab+8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^2b^2\left(a^2b^2-1\right)+8\left(ab-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(ab-1\right)\left[a^2b^2\left(ab+1\right)+8\right]=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab-1\end{matrix}\right.\left(a,b>0\right)\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}=1\\\sqrt{\dfrac{x+y}{y-1}}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=x+2\\x+y=y-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
