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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
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)
\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)
\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)
\(=0\)
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Đặt \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=k\) \(\left(k\ne0\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=a.k\\y=b.k\\z=c.k\end{matrix}\right.\)
Ta có :
\(A=\dfrac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
\(A=\dfrac{\left[\left(a.k\right)^2+\left(b.k\right)^2+\left(c.k\right)^2\right]\cdot\left(a^2+b^2+c^2\right)}{\left(a.a.k+b.b.k+c.c.k\right)^2}\)
\(A=\dfrac{\left(a^2k^2+b^2k^2+c^2k^2\right)\cdot\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(A=1\)
Bài 2: Restore : a;b;c không âm thỏa \(a^2+b^2+c^2=1\)
Tìm Min & Max của \(M=\left(a+b+c\right)^3+a\left(2bc-1\right)+b\left(2ac-1\right)+c\left(2ab-1\right)\)
Bài 4: Tương đương giống hôm nọ thôi : V
Bài 5 : Thiếu ĐK thì vứt luôn : V
Bài 7: Tương đương
( Hoặc có thể AM-GM khử căn , sau đó đổi \(\left(a;b;c\right)\rightarrow\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\) rồi áp dụng bổ đề vasile)
Bài 8 : Đây là 1 dạng của BĐT hoán vị
@Ace Legona @Akai Haruma @Hung nguyen @Hà Nam Phan Đình @Neet
\(A=\dfrac{n!+2}{\dfrac{n!}{\left(n-k\right)!}\cdot n!-k}+\dfrac{3003+10010+6435}{19448}\)
\(=\dfrac{n!+2}{n\left(n-1\right)\cdot...\cdot\left(n-k+1\right)\cdot n!-k}+1=\dfrac{n!+2+\dfrac{n!^2}{\left(n-k\right)!}-k}{\dfrac{n!^2}{\left(n-k\right)!}-k}\)
\(B=\dfrac{n!-\left(n-1\right)!}{\left(n-2\right)!}=\dfrac{\left(n-1\right)!\left(n-1\right)}{\left(n-2\right)!}=\left(n-1\right)^2=n^2-2n+1\)