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\(P=\frac{1}{25a}+\frac{1}{16b}+\frac{1}{9c}=\frac{\frac{1}{25}}{a}+\frac{\frac{1}{16}}{b}+\frac{\frac{1}{9}}{c}\ge\frac{\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)^2}{a+b+c}=\frac{2209}{3600}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{\frac{1}{5}}{a}=\frac{\frac{1}{4}}{b}=\frac{\frac{1}{3}}{c}=\frac{\frac{1}{5}+\frac{1}{4}+\frac{1}{3}}{a+b+c}=\frac{47}{60}\)
\(\Rightarrow\)\(\hept{\begin{cases}a=\frac{1}{5}:\frac{47}{60}=\frac{12}{47}\\b=\frac{1}{4}:\frac{47}{60}=\frac{15}{47}\\c=\frac{1}{3}:\frac{47}{60}=\frac{20}{47}\end{cases}}\)
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Ta có: \(a< a+b\left(a,b>0\right)\Rightarrow\frac{a}{a+b}< 1\)
Có: \(\frac{a}{a+b}=\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}\)
Lại có: \(\frac{a}{b+a}< 1\Leftrightarrow\sqrt{\frac{a}{b+a}}< 1\Rightarrow\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+b}}< \sqrt{\frac{a}{a+b}}\Rightarrow\frac{a}{a+b}< \sqrt{\frac{a}{a+b}}\)
Chứng minh tương tự ta có:
\(\frac{b}{b+c}< \sqrt{\frac{b}{b+c}}\)
\(\frac{c}{c+a}< \sqrt{\frac{c}{c+a}}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\)
đpcm
Sai thì thôi nhé~
Mới lp 8
Easy nà!
Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\) thì xyz = 1
BĐT trở thành: \(x^2+y^2+z^2\ge x+y+z\)
Áp dụng BĐT AM-GM,ta có: \(VT+1=\left(x^2+y^2\right)+\left(z^2+1\right)\)
\(\ge2xy+2z\ge2\sqrt{2xy.2z}=4\sqrt{xyz}=4\)
Suy ra \(VT\ge3\) (1)
Lại có: \(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)
Cộng theo vế 3 BĐT: \(VT+3\ge2\left(x+y+z\right)\)
Kết hợp (1) suy ra \(2VT\ge VT+3\ge2\left(x+y+z\right)=2VP\)
Từ đây,ta có:\(2VT\ge2VP\Rightarrow VT\ge VP^{\left(đpcm\right)}\)
Dấu "=" xảy ra khi x = y = z = 1
\(\frac{xy}{z}+\frac{yz}{x}\ge2y\) ; \(\frac{xy}{z}+\frac{zx}{y}\ge2x\); \(\frac{yz}{x}+\frac{zx}{y}\ge2z\)
Cộng vế với vế:
\(2\left(\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\right)\ge2\left(x+y+z\right)\)
Dấu "=" xảy ra khi \(x=y=z\)
Áp dụng BĐT cô si với hai số không âm, Ta có:
\(\left(a+b+c\right)^2=1\ge4a\left(b+c\right)\)
\(\Leftrightarrow b+c\ge4a\left(b+c\right)^2\)
Mà \(\left(b+c\right)^2\ge4bc\forall b,c\ge0\)
\(\Rightarrow b+c\ge16abc\)
Dấu "=" xảy ra khi:
\(\hept{\begin{cases}a+b+c=1\\b=c\\a=b+c\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{1}{2}\\b=c=\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)
Áp dụng BĐT Cô si với 2 số dương ta có:
\(\frac{a}{b}+\frac{b}{a}\ge2,\frac{b}{c}+\frac{c}{b}\ge2,\frac{c}{a}+\frac{a}{c}\ge2\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge6\)(đúng)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)(do a+b+c=1)
\(VT=\frac{25a}{b+c}+25+\frac{16b}{a+c}+16+\frac{c}{a+b}+1-42\)
\(VT=\frac{25\left(a+b+c\right)}{b+c}+\frac{16\left(a+b+c\right)}{a+c}+\frac{a+b+c}{a+b}-42\)
\(VT=\left(a+b+c\right)\left(\frac{25}{b+c}+\frac{16}{a+c}+\frac{1}{a+b}\right)-42\)
\(VT\ge\left(a+b+c\right).\frac{\left(5+4+1\right)^2}{b+c+a+c+a+b}-42=\frac{100\left(a+b+c\right)}{2\left(a+b+c\right)}-42=8\)
Dấu "=" xảy ra khi: \(\frac{b+c}{5}=\frac{a+c}{4}=\frac{a+b}{1}=\frac{2\left(a+b+c\right)}{5+4+1}=\frac{a+b+c}{5}\)
\(\Rightarrow a=0\) trái giả thiết a dương, vậy dấu "=" không xảy ra
\(\Rightarrow\frac{25a}{b+c}+\frac{16b}{a+c}+\frac{c}{a+b}>8\)
Cho mik hỏi chỗ dòng thứ tư là bạn sd BĐT nào đc k ạ