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Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
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
\(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-15}{z}\)
Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)
\(B=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}+3\) (1)
Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)
\(\Leftrightarrow xy+yz\ge y^2+zx\)
\(\Leftrightarrow\dfrac{x}{z}+1\ge\dfrac{y}{z}+\dfrac{x}{y}\)
Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)
Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)
Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)
Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)
\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)
\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)
\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)
Đành giải tạm bằng nick này vì sợ một vài thành phần trẻ trâu anti phá phách :poor:
Phân tích và giải
Dễ thấy: Dấu "=" khi \(a=b=c=1\)
\(\Rightarrow L=Σ\dfrac{a}{\left(a+1\right)^2}=\dfrac{3}{4}\text{ và }F=-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=-\dfrac{1}{2}\)
Khi đó \(VT=L-F=\dfrac{3}{4}-\dfrac{1}{2}=\dfrac{1}{4}\)
Ta sẽ chia làm 2 bước cm:
B1: \(Σ\dfrac{a}{\left(a+1\right)^2}\le\dfrac{3}{4}\). Ta xét BĐT :
\(\dfrac{a}{\left(a+1\right)^2}=\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2k}+a^k\right)}{8\left(a^{2k}+a^k+1\right)}\) (cần tìm \(k\) thỏa mãn)
\(\Leftrightarrow8a\left(a^{2k}+a^k+1\right)-3\left(a^{2k}+a^k\right)\left(a^2+2a+1\right)\le0\)\(\Leftrightarrow f\left(a\right)=-3a^{2k}+2a^{k+1}-3a^{k+2}+2a^{2k+1}-3a^{2k+2}-3a^k+8a\)
\(\Rightarrow f'\left(a\right)=2k\cdot-3a^{2k-1}+\left(k+1\right)2a^k-\left(k+2\right)3a^{k+1}+\left(2k+1\right)2a^{2k}-\left(2k+2\right)3a^{2k+1}-k\cdot3a^{k-1}+8a\)
\(\Rightarrow f'\left(1\right)=0\Rightarrow-12k=0\Rightarrow k=0\)
Hay BĐT phụ cần tìm là \(\dfrac{a}{a^2+2a+1}\le\dfrac{3\left(a^{2\cdot0}+a^0\right)}{8\left(a^{2\cdot0}+a^0+1\right)}=\dfrac{1}{4}\) (bài này \(k\) đẹp ra luôn \(\farac{1}{4}\) cộng vào là ok =))
\(\Leftrightarrow-\dfrac{\left(a-1\right)^2}{4\left(a+1\right)^2}\le0\) *Đúng* \(\RightarrowΣ\dfrac{a}{\left(a+1\right)^2}\leΣ\dfrac{1}{4}=\dfrac{3}{4}\)
B2: CM \(-\dfrac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\le-\dfrac{1}{2}\)
Tự cm nhé Goodluck :v
câu 1: \(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Ta có:
\(a\left(b+c\right)^2+b\left(c+a\right)^2+c\left(a+b\right)^2=4abc\)
\(\Leftrightarrow\left(ab+ac\right)\left(b+c\right)+b\left(c^2+2ac+a^2\right)+c\left(a^2+2ab+b^2\right)=4abc\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc^2+2abc+ba^2+ca^2+2abc+cb^2-4abc=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+\left(bc^2+cb^2\right)+\left(ba^2+ca^2\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc\left(b+c\right)+a^2\left(b+c\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac+bc+a^2\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left[b\left(c+a\right)+a\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}b=-c\\a=-b\\c=-a\end{matrix}\right.\)
Ta lại có:
\(a^{2013}+b^{2013}+c^{2013}=1\)
Với : \(b=-c\Leftrightarrow a^{2013}-c^{2013}+c^{2013}=1\Leftrightarrow a=1\)
\(\Rightarrow M=\dfrac{1}{a^{2015}}+\dfrac{1}{b^{2015}}+\dfrac{1}{c^{2015}}=\dfrac{1}{1}+\dfrac{-1}{c^{2015}}+\dfrac{1}{c^{2015}}=1\)
Mà do \(a,b,c\) bình đẳng nên với trường hợp nào đều là \(M=1\)