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d/ Đặt \(x=a+b\) , \(y=b+c\) , \(z=c+a\)
thì : \(a=\frac{x+z-y}{2}\) ; \(b=\frac{x+y-z}{2}\) ; \(c=\frac{y+z-x}{2}\)
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}+\frac{\frac{y+z-x}{2}}{x}\)
\(=\frac{z+x-y}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{z}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}-3\right)\)
\(=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)-\frac{3}{2}\ge\frac{1}{2}.6-\frac{3}{2}=\frac{3}{2}\)
b/ \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)
\(\Leftrightarrow\left(a^2b^2-2abc+c^2\right)+\left(b^2c^2-2abc+a^2\right)+\left(c^2a^2-2abc+b^2\right)\ge0\)
\(\Leftrightarrow\left(ab-c\right)^2+\left(bc-a\right)^2+\left(ca-b\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu dc chứng minh.
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
a/ Từ BĐT ban đầu ta có:
\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) (đpcm)
b/ Chia 2 vế của BĐT ở câu a cho 9 ta được:
\(\frac{a^2+b^2+c^2}{3}\ge\frac{\left(a+b+c\right)^2}{9}=\left(\frac{a+b+c}{3}\right)^2\) (đpcm)
c/ Cộng 2 vế của BĐT ban đầu với \(2ab+2bc+2ca\) ta được:
\(a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d/ Áp dụng BĐT ban đầu cho các số \(a^2;b^2;c^2\) ta được:
\(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\ge a^2b^2+b^2c^2+c^2a^2\)
Mặt khác ta cũng có:
\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge ab.bc+bc.ca+ab+ca=abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
a)Áp dụng Bđt Cô si ta có:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{3}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\ge\frac{3\sqrt[3]{abc}}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Cộng theo vế 2 bđt trên ta có:
\(3\ge\frac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
Dấu = khi a=b=c
b)Áp dụng Bđt Cô-si ta có:
\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc^2a}{ab}}=2c\)
\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca^2b}{bc}}=2a\)
\(\frac{bc}{a}+\frac{ab}{c}\ge2\sqrt{\frac{b^2ac}{ac}}=2b\)
Cộng theo vế 3 bđt trên ta có:
\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)
Đấu = khí a=b=c
Áp dụng bđt AM-GM:
\(\frac{1}{a^3\left(b+c\right)}+\frac{a\left(b+c\right)}{4}\ge2\sqrt{\frac{a\left(b+c\right)}{4a^3\left(b+c\right)}}=\frac{1}{a}\)
\(\frac{1}{b^3\left(c+a\right)}+\frac{b\left(c+a\right)}{4}\ge2\sqrt{\frac{b\left(c+a\right)}{4b^3\left(c+a\right)}}=\frac{1}{b}\)
\(\frac{1}{c^3\left(a+b\right)}+\frac{c\left(a+b\right)}{4}\ge2\sqrt{\frac{c\left(a+b\right)}{4c^3\left(a+b\right)}}=\frac{1}{c}\)
Cộng theo vế:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}+\frac{ab+bc+ac}{2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{ab+bc+ac}{2}\)
\(\Leftrightarrow\frac{2}{a^3\left(b+c\right)}+\frac{2}{b^3\left(c+a\right)}+\frac{2}{c^3\left(a+b\right)}\ge ab+bc+ac\) (đpcm)
\("="\Leftrightarrow a=b=c=1\)
a.
\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
(luôn đúng)
b. Áp dụng BĐT \(x^2+y^2\ge2xy\)
\(a^2+b^2\ge2ab,a^2+1\ge2a,b^2+1\ge2b\)\(\Rightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\Leftrightarrow a^2+b^2+1\ge ab+a+b\)
c. Tương tự câu b
Áp dụng BĐT Cô si ta có
i. \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},\frac{1}{b}+\frac{1}{c}\ge\frac{2}{\sqrt{bc}},\frac{1}{c}+\frac{1}{a}\ge\frac{2}{\sqrt{ca}}\)
\(\Rightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\)
k. Tương tự câu i
\(\left(a+\frac{4b}{c^2}\right)\left(b+\frac{4c}{a^2}\right)\left(c+\frac{4a}{b^2}\right)\ge2\sqrt{\frac{4ab}{c^2}}.2\sqrt{\frac{4bc}{a^2}}.2\sqrt{\frac{4ac}{b^2}}=64\)
Dấu "=" xảy ra khi \(a=b=c=2\)
\(\frac{a^3}{b}+ab\ge2a^2\) ; \(\frac{b^3}{c}+bc\ge2b^2\); \(\frac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\ge2\left(ab+bc+ca\right)-\left(ab+bc+ca\right)=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{b^2c^2}{ab+ca}+\frac{c^2a^2}{bc+ab}+\frac{a^2b^2}{ca+bc}\)
\(\ge\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{1}{2}\left(ab+bc+ca\right)\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel(hoặc áp dụng BĐT quen thuộc: \(\frac{p^2}{m}+\frac{q^2}{n}\ge\frac{\left(p+q\right)^2}{m+n}\) 2 lần),ta có:
\(VT=\frac{\left(\frac{1}{a^2}\right)}{a\left(b+c\right)}+\frac{\left(\frac{1}{b^2}\right)}{b\left(c+a\right)}+\frac{\left(\frac{1}{c^2}\right)}{c\left(a+b\right)}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}\) (thay abc = 1 vào)
\(=\frac{ab+bc+ca}{2}=\frac{1}{2}\left(ab+bc+ca\right)^{\left(đpcm\right)}\)