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1. Đề thiếu
2. BĐT cần chứng minh tương đương:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Ta có:
\(a^4+b^4+c^4\ge\dfrac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\dfrac{1}{3}\left(ab+bc+ca\right)^2\ge\dfrac{1}{3}.3abc\left(a+b+c\right)\) (đpcm)
3.
Ta có:
\(\left(a^6+b^6+1\right)\left(1+1+1\right)\ge\left(a^3+b^3+1\right)^2\)
\(\Rightarrow VT\ge\dfrac{1}{\sqrt{3}}\left(a^3+b^3+1+b^3+c^3+1+c^3+a^3+1\right)\)
\(VT\ge\sqrt{3}+\dfrac{2}{\sqrt{3}}\left(a^3+b^3+c^3\right)\)
Lại có:
\(a^3+b^3+1\ge3ab\) ; \(b^3+c^3+1\ge3bc\) ; \(c^3+a^3+1\ge3ca\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(ab+bc+ca\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow VT\ge\sqrt{3}+\dfrac{6}{\sqrt{3}}=3\sqrt{3}\)
4.
Ta có:
\(a^3+1+1\ge3a\) ; \(b^3+1+1\ge3b\) ; \(c^3+1+1\ge3c\)
\(\Rightarrow a^3+b^3+c^3+6\ge3\left(a+b+c\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
5.
Ta có:
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\) ; \(\dfrac{a}{b}+\dfrac{c}{a}\ge2\sqrt{\dfrac{c}{b}}\) ; \(\dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\)
\(\Rightarrow\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}+\sqrt{\dfrac{a}{c}}\le\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=1\)
\(sigma\frac{a}{1+b-a}=sigma\frac{a^2}{a+ab-a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\frac{1}{b^2+c^2}=\frac{1}{1-a^2}=1+\frac{a^2}{b^2+c^2}\le1+\frac{a^2}{2bc}\)
Tương tự cộng lại quy đồng ta có đpcm
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Bunhiacopxki:
\(\left(a^2+b+c+d\right)\left(1+b+c+d\right)\ge\left(a+b+c+d\right)^2=16\)
\(\Rightarrow\dfrac{1}{a^2+b+c+d}\le\dfrac{1+b+c+d}{16}\)
Tương tự:
\(\dfrac{1}{b^2+c+d+a}\le\dfrac{1+c+d+a}{16}\) ; \(\dfrac{1}{c^2+d+a+b}\le\dfrac{1+d+a+b}{16}\)
\(\dfrac{1}{d^2+a+b+c}\le\dfrac{1+a+b+c}{16}\)
Cộng vế:
\(P\le\dfrac{4+3\left(a+b+c+d\right)}{16}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d=1\)
Bài toán cơ bản:
\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)
Bunhiacopxki:
\(\left(a+b+c\right)\left(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right)\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\right)^2=1\)
\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)
BĐT cần c/m trở thành:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)
\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)
\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)
\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)
Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)
Nên (1) tương đương:
\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)
\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)
BĐT trên hiển nhiên đúng theo AM-GM do:
\(\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\sqrt[6]{\dfrac{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}{abc\left(1-a\right)\left(1-b\right)\left(1-c\right)}}=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)
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Ta có:
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(a+b+c\right)^2=9\)
\(\Rightarrow\dfrac{1}{b^2+c^2+1}\le\dfrac{a^2+2}{9}\)
\(\Rightarrow\dfrac{a}{b^2+c^2+1}\le\dfrac{a^3+2a}{9}\)
Tương tự: \(\dfrac{b}{c^2+a^2+1}\le\dfrac{b^3+2b}{9}\) ; \(\dfrac{c}{a^2+b^2+1}\le\dfrac{c^3+2c}{9}\)
Cộng vế:
\(VT\le\dfrac{a^3+b^3+c^3+2\left(a+b+c\right)}{9}=\dfrac{a^3+b^3+c^3+6}{9}\) (1)
Lại có:
\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)
\(\Rightarrow a^3+b^3+c^3\ge3\Rightarrow6\le2\left(a^3+b^3+c^3\right)\) (2)
(1);(2) \(\Rightarrow VT\le\dfrac{a^3+b^3+c^3+2\left(a^3+b^3+c^3\right)}{9}=\dfrac{a^3+b^3+c^3}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Đề là
Cho \(a;b;c\ge0\) thỏa mãn a+b+c = 1
Cmr : \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\ge\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}\) ak bạn
Ta có:a+b+c=1
\(đpcm\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{2}{a+2b+c}+\frac{2}{2a+b+c}+\frac{2}{a+b+2c}\)(*)
Áp dụng BĐT Bunhiacopxki:
\(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)(1)
Tương tự:\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{a+b+2c}\)(2)
\(\frac{1}{a+b}+\frac{1}{c+a}\ge\frac{4}{2a+b+c}\)(3)
Cộng theo từng vế của (1);(2);(3) ta đc:(*)(đpcm)
Dấu ''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)