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3: =>a^3+b^3+c^3>=3abc
=>(a+b)^3+c^3-3ab(a+b)-3abc>=0
=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)>=0
=>a^2+b^2+c^2-ab-bc-ac>=0
=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0
=>(a-b)^2+(a-c)^2+(b-c)^2>=0(luôn đúng)
Có:
\(\left\{{}\begin{matrix}a^3+b^3\ge ab\left(a+b\right)\\b^3+c^3\ge bc\left(b+c\right)\\c^3+a^3\ge ca\left(c+a\right)\end{matrix}\right.\Rightarrow VT\ge a^2.\frac{b+c}{2}+b^2.\frac{c+a}{2}+c^2.\frac{a+b}{2}\ge a^2\sqrt{bc}+b^2\sqrt{ca}+c^2\sqrt{ab}=VP\)
Dấu bằng xảy ra khi a=b=c
3: =>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2
=>a^2d^2-2abcd+b^2c^2>=0
=>(ad-bc)^2>=0(luôn đúng)
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
\((a^2+2c^2)(1+2)\geq (a+2c)^2\)
\(\Rightarrow \sqrt{a^2+2c^2}\geq \frac{a+2c}{\sqrt{3}}\)
\(\Rightarrow \frac{\sqrt{a^2+2c^2}}{ac}\geq \frac{a+2c}{\sqrt{3}ac}=\frac{ab+2bc}{\sqrt{3}abc}\)
Hoàn toàn tương tự: \(\left\{\begin{matrix} \frac{\sqrt{c^2+2b^2}}{bc}\geq \frac{ac+2ab}{\sqrt{3}abc}\\ \frac{\sqrt{b^2+2a^2}}{ab}\geq \frac{bc+2ac}{\sqrt{3}abc}\end{matrix}\right.\)
Cộng theo vế các BĐT trên thu được:
\(\text{VT}\geq \frac{1}{\sqrt{3}}.\frac{ab+2bc+ac+2ab+bc+2ac}{abc}=\frac{1}{\sqrt{3}}.\frac{3(ab+bc+ac)}{abc}=\frac{1}{\sqrt{3}}.\frac{3abc}{abc}=\sqrt{3}\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c=3$
Bài 2: Bài này sử dụng pp xác định điểm rơi thôi.
Áp dụng BĐT AM-GM ta có:
\(24a^2+24.(\frac{31}{261})^2\geq 2\sqrt{24^2.(\frac{31}{261})^2a^2}=\frac{496}{87}a\)
\(b^2+(\frac{248}{87})^2\geq 2\sqrt{(\frac{248}{87})^2.b^2}=\frac{496}{87}b\)
\(93c^2+93.(\frac{8}{261})^2\geq 2\sqrt{93^2.(\frac{8}{261})^2c^2}=\frac{496}{87}c\)
Cộng theo vế:
\(B+\frac{248}{29}\geq \frac{496}{87}(a+b+c)=\frac{496}{87}.3=\frac{496}{29}\)
\(\Rightarrow B\geq \frac{496}{29}-\frac{248}{29}=\frac{248}{29}\)
Vậy \(B_{\min}=\frac{248}{29}\). Dấu bằng xảy ra khi: \((a,b,c)=(\frac{31}{261}; \frac{248}{87}; \frac{8}{261})\)
Câu 2)
Ta có \(\frac{1}{a+1}+\frac{1}{b+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{b+1+a+1}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{\left(a+1\right)\left(b+1\right)}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{\left(a+1\right)b+a+1}\ge\frac{4}{3}\)
\(\Rightarrow\frac{3}{ab+b+a+1}\ge\frac{4}{3}\)
Ta có \(a+b=1\)
\(\Rightarrow\frac{3}{ab+2}\ge\frac{4}{3}\)
\(\Leftrightarrow9\ge4\left(ab+2\right)\)
\(\Rightarrow9\ge4ab+8\)
\(\Rightarrow1\ge4ab\)
Do \(a+b=1\Rightarrow\left(a+b\right)^2=1\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
\(\Rightarrow a^2+2ab+b^2\ge4ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) (đpcm )
Câu 3)
Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)
Mà \(a+b+c=1\)
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge9\)
\(\Rightarrow a+b+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Áp dụng bất đẳng thức Cô-si
\(\Rightarrow\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc}\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (điều này luôn luôn đúng)
\(\Rightarrow\) ĐPCM
\(P=\frac{b^2c^2+c^2a^2+a^2b^2}{abc}\Rightarrow P^2=\frac{b^4c^4+c^4a^4+a^4b^4+2a^2b^2c^2\left(a^2+b^2+c^2\right)}{a^2b^2c^2}\)
\(P^2\ge\frac{a^2b^2c^2\left(a^2+b^2+c^2\right)+2a^2b^2c^2}{a^2b^2c^2}=\frac{3a^2b^2c^2}{a^2b^2c^2}=3\)
\(\Rightarrow P\ge\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
ta co: (ab+bc+ac)2 - 3abc(a+b+c) = a2b2+ b2c2 + a2c2 + 2a2bc + 2b2ac+ 2c2ab- 3a2bc- 3b2ac- 3c2ab.
=a2b2+ b2c2 + a2c2- a2bc- b2ac-c2ab.
=>cm: a2b2+ b2c2 + a2c2- a2bc- b2ac- c2ab >= 0
<=> 2(a2b2+ b2c2 + a2c2- a2bc- b2ac- c2ab) >=0
<=> (ab- ac)2 + (ab- bc)2 + (bc- ac)2 >=0 (luon dung voi moi a,b,c)
=> dpcm.