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\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2\ge5\sqrt[5]{\dfrac{a^{20}b^2}{b^{12}}}=5.\dfrac{a^4}{b^2}\)
\(\Rightarrow4.\dfrac{a^5}{b^3}+b^2\ge5.\dfrac{a^4}{b^2}\)
Tương tự: \(4.\dfrac{b^5}{c^3}+c^2\ge5\dfrac{b^4}{c^2};4\dfrac{c^5}{a^3}+a^2\ge5.\dfrac{c^4}{a^2}\)
\(\Rightarrow4\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)
Lại có: \(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5a^2\)
\(\Rightarrow2.\dfrac{a^5}{b^3}+3b^2\ge5a^2\), tương tự: \(2.\dfrac{b^5}{c^3}+3c^2\ge5b^2;2\dfrac{c^5}{a^3}+3a^2\ge5c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}+4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge4.\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+a^2+b^2+c^2\ge5.\left(\dfrac{c^4}{a^2}+\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}\right)\)
\(\Rightarrow dpcm\)
giả sử \(a>b>c>0\) thì ta có :
\(\dfrac{a^4}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^4}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^4}{a^2}\left(\dfrac{c}{a}-1\right)\ge\dfrac{2a^2b}{c}+\dfrac{c^5}{a^3}-\dfrac{c^4}{a^2}\)
\(\ge\dfrac{2c^4b}{a}-\dfrac{c^4}{a^2}=\dfrac{c^4}{a}\left(2b-\dfrac{1}{a}\right)>0\)
làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
mấy câu cậu câu đăng khác bn làm tương tự nha . nếu bn lm không được thì có j mk lm luôn cho còn h mk bạn rồi :(
Ta có:
\(\sum\left(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\right)\ge5\sum a^2\)
\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(a^2+b^2+c^2\right)-3\left(a^2+b^2+c^2\right)=2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
Ap dung BDT Cauchy-Schwarz ta co:
\(\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\left(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\right)\ge\left(a^2+b^2+c^2\right)^2\)
Can chung minh \(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\ge a^2+b^2+c^2\)
\(VT=\dfrac{a^4}{ac}+\dfrac{b^4}{ab}+\dfrac{c^4}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{ab+bc+ca}=a^2+b^2+c^2=VP\)
\("="\Leftrightarrow a=b=c\)
Từ bài toán này (mà bạn đã hỏi cách đây vài bữa):
cho a,b,c>0. Chứng minh rằng: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\) - Hoc24
Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b+c}{\sqrt[3]{abc}}\)
Do đó: \(VT\ge\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\)
Lại có: \(\dfrac{a+b+c}{\sqrt[3]{abc}}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}=3\)
Đặt \(\dfrac{a+b+c}{\sqrt[3]{abc}}=x\ge3\Rightarrow VT\ge x+\dfrac{1}{x}=\dfrac{x}{9}+\dfrac{1}{x}+\dfrac{8x}{9}\ge2\sqrt{\dfrac{x}{9x}}+\dfrac{8}{9}.3=\dfrac{10}{3}\) (đpcm)
\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)
\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
giả sử \(a>b>c>0\) thì ta có :
\(\dfrac{a^2}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^2}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\ge2\dfrac{a}{b}+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\)
\(=\dfrac{2a}{b}+\dfrac{c^3}{a^3}-\dfrac{c^2}{a^2}\ge0\)
làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)
tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)
suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
suy ra dpcm
dau = xay ra khi a=b=c
Ta có \(\dfrac{a^3+b^3}{2ab}\ge\dfrac{ab\left(a+b\right)}{2ab}=\dfrac{a+b}{2}\)
(áp dụng BĐT quen thuộc \(a^3+b^3\ge ab\left(a+b\right)\))
Lập 2 BĐT tương tự rồi cộng theo vế:
\(VT\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có đpcm.
Áp dụng BĐT \(AM-GM\) ta có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{15}b^4}{b^9}}=5\dfrac{a^3}{b}\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{15}c^4}{c^9}}=5\dfrac{b^3}{c}\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{15}a^4}{a^9}}=5\dfrac{c^3}{a}\)
Cộng từng vế của BĐT ta được :
\(3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
Tiếp tục áp dụng BĐT \(AM-GM\) ta lại có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{10}b^6}{b^6}}=5a^2\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{10}c^6}{c^6}}=5b^2\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{10}a^6}{a^6}}=5c^2\)
Cộng vế theo vế ta được :
\(2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+3\left(a^2+b^2+c^2\right)\ge5\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow5\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\left(đpcm\right)\)
Bạn có cách nào ko đụng AM- GM 5 số không ( chứng minh chắc chết ) . Thầy mình gợi ý dùng bđt phụ a^3 + b^3 >= ab(a+b)