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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 :(
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 :
\(2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+3\left(a^2+b^2+c^2\right)\ge5\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow5\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\left(đpcm\right)\)
\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)
\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
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
2, a, \(a+\dfrac{1}{a}\ge2\)
\(\Leftrightarrow\dfrac{a^2+1}{a}\ge2\)
\(\Rightarrow a^2-2a+1\ge0\left(a>0\right)\)
\(\Leftrightarrow\left(a-1\right)^2\ge0\)( là đt đúng vs mọi a)
vậy...................
Câu 1:
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}=3\)
\(M=\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{5-\sqrt{5}+1}=\sqrt{6-\sqrt{5}}\)
\(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\)
\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}\)
\(=\dfrac{a^2+b^2+c^2}{2}=\dfrac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{\sqrt{3}}\)
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
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 \(\dfrac{b^3}{a}+\dfrac{c^3}{b}+\dfrac{a^3}{c}\ge a^2+b^2+c^2\)
\(VT=\dfrac{a^4}{ac}+\dfrac{b^4}{ab}+\dfrac{c^4}{bc}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)}{ab+bc+ca}=a^2+b^2+c^2=VP\)
\("="\Leftrightarrow a=b=c\)