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Chứng minh BĐT Phụ: \(a^5+b^5\ge a^4b+ab^4\)với \(a;b>0\)
\(\Rightarrow\frac{a^5+b^5}{ab\left(a+b\right)}\ge\frac{a^4b+ab^4}{ab\left(a+b\right)}=\frac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\frac{ab\left(a+b\right)\left(a^2-ab+b^2\right)}{ab\left(a+b\right)}=a^2-ab+b^2\)
Áp dụng ta có: \(VT\)(VẾ TRÁI)\(\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\) \(\left(1\right)\)
Xét: \(\left[2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\right]-\left[3\left(ab+bc+ca\right)-2\right]\)
\(=2\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)+2\)
\(=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\) (Do a2+b2+c2=1) \(\left(2\right)\)
Mà \(a^2+b^2+c^2\ge ab+bc+ca\) Tự chứng minh \(\left(3\right)\)
Từ (1);(2) và (3) suy ra \(VT\ge3\left(ab+bc+ca\right)-2\)
Vậy \(\frac{a^5+b^5}{ab\left(a+b\right)}+\frac{b^5+c^5}{bc\left(b+c\right)}+\frac{c^5+a^5}{ca\left(c+a\right)}\ge3\left(ab+bc+ca\right)-2\)
Ta có \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)\(\Rightarrow3\sqrt[3]{a^2b^2c^2}\le3\Leftrightarrow abc\le1\)
\(\Rightarrow\)\(\frac{1}{1+a^2\left(b+c\right)}\le\frac{1}{abc+a^2\left(b+c\right)}\)\(=\frac{1}{a\left(ab+bc+ca\right)}=\frac{1}{3a}\)
\(CMTT\Rightarrow\frac{1}{1+b^2\left(c+a\right)}\le\frac{1}{3b}\)
\(\frac{1}{1+c^2\left(a+b\right)}\le\frac{1}{3c}\)
\(\Rightarrow VT\le\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}\)\(=\frac{ab+bc+ca}{3abc}=\frac{1}{abc}\)
\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
...
\(sigma\frac{a^2+b^2}{ab\left(a+b\right)^3}\ge sigma\frac{\frac{\left(a+b\right)^2}{2}}{\left(a+b\right)^2\left(a^3+b^3\right)}=sigma\frac{1}{2\left(a^3+b^3\right)}\ge\frac{9}{4\left(a^3+b^3+c^3\right)}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
Câu hỏi của Lê Minh Đức - Toán lớp 9 - Học toán với OnlineMath
Đây nha! Vô tcn xem ảnh!
Lời giải:
Vì $abc=1$ nên:
\((a+bc)(b+ac)(c+ab)=a(a+bc)b(b+ac)c(c+ab)=(a^2+1)(b^2+1)(c^2+1)\)
Áp dụng BĐT Bunhiacopxky:
\((a^2+1)(1+b^2)\geq (a+b)^2; (a^2+1)(1+c^2)\geq (a+c)^2; (b^2+1)(1+c^2)\geq (b+c)^2\)
Nhân theo vế và thu gọn:
\(\Rightarrow (a^2+1)(b^2+1)(c^2+1)\geq (a+b)(b+c)(c+a)\)
Lại có: Theo BĐT AM-GM thì:
\((a+b)(b+c)(c+a)=(ab+bc+ac)(a+b+c)-abc\)
\(\geq (ab+bc+ac)(a+b+c)-\frac{(a+b+c)(ab+bc+ac)}{9}=\frac{8(a+b+c)(ab+bc+ac)}{9}(*)\) (đây là BĐT khá quen thuộc rồi)
Do đó:
\(P=\frac{(a+bc)(b+ca)(c+ab)}{ab+bc+ac}+\frac{1}{a+b+c}=\frac{(a^2+1)(b^2+1)(c^2+1)}{ab+bc+ac}+\frac{1}{a+b+c}\geq \frac{(a+b)(b+c)(c+a)}{ab+bc+ac}+\frac{1}{a+b+c}\)
\(P\geq \frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\)
Áp dụng BĐT (*) và AM-GM:
\(\frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}\geq 7.\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(ab+bc+ac)}=\frac{7}{9}(a+b+c)\geq \frac{7}{9}.3\sqrt[3]{abc}=\frac{7}{3}\)
\(\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\geq 2\sqrt{\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)(a+b+c)}}\geq 2\sqrt{\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(a+b+c)(ab+bc+ac)}}=\frac{2}{3}\)
\(\Rightarrow P\geq \frac{7}{3}+\frac{2}{3}=3\)
Vậy $P_{\min}=3$
\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\)
\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1\)
\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1+1-1\)
Áp dụng BĐT AM-GM ta có:
\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\ge a^2+b^2+c^2+2ab+2bc+2ac-1=\left(a+b+c\right)^2-1\)\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\)
Dấu " = " xảy ra <=> ...
Ta có: \(\frac{1}{3}.\left(a+b+c\right)^2\ge ab+bc+ca\)( BĐT quen thuộc tự c/m)
\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\ge\frac{\left(a+b+c\right)^2}{\frac{1}{3}\left(a+b+c\right)^2}-\frac{1}{\frac{1}{3}\left(a+b+c\right)}+\frac{1}{a+b+c}\)\(=3+\frac{a+b+c-3}{\left(a+b+c\right)^2}\)
Ta có: \(abc=1\Leftrightarrow\sqrt[3]{abc}=1\le\frac{a+b+c}{3}\left(AM-GM\right)\)
\(\Rightarrow a+b+c\ge3\)
Dấu " = " xảy ra <=> ...
\(\Rightarrow P\ge3+\frac{a+b+c-3}{\left(a+b+c\right)^2}\ge3\)
Dấu " = " xảy ra <=> a=b=c=1
KL:...........
\(\frac{\left(a+b\right)^2}{ab}+\frac{\left(b+c\right)^2}{bc}+\frac{\left(c+a\right)^2}{ac}=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}+6\)
\(bđt\Leftrightarrow\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge3+2\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)\)
Mà: \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)\ge\frac{4a}{b+c}+\frac{4b}{a+c}+\frac{4c}{a+b}\)
\(\Leftrightarrow2\left(\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\right)\ge3\Leftrightarrow\frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+c}\ge\frac{3}{2}\)
bđt cuối đúng theo Nesbit. Dấu "=" xảy ra khi a=b=c
Ta co:
\(\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\)
\(=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\left(ab+bc+ca\right)\le\text{ }\frac{\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)\right]^3}{27}\)
\(\frac{\left[a^2+b^2+c^2+2\left(ab+bc+ca\right)\right]^3}{27}=\frac{\left(a+b+c\right)^6}{27}=\frac{3^6}{27}=27\)
Dau '=' xay ra khi \(a=b=c=1\)