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\(VT=\left(\dfrac{b}{a}+\dfrac{b}{c}\right)+\left(\dfrac{c}{a}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{a}{c}\right)\)
Ta có \(\left(\dfrac{b}{c}+\dfrac{b}{a}\right)\left(a+c\right)\ge\left(\sqrt{b}+\sqrt{b}\right)^2=4b\Leftrightarrow\dfrac{b}{c}+\dfrac{b}{a}\ge\dfrac{4b}{a+c}\)
CMTT \(\Leftrightarrow\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge\dfrac{4c}{a+b};\dfrac{a}{b}+\dfrac{a}{c}\ge\dfrac{4a}{b+c}\)
Cộng VTV ta đc đpcm
Dấu \("="\Leftrightarrow a=b=c\)
\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\dfrac{9}{4}\)
\(\Rightarrow\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)
Dấu "=" xảy ra khi \(a=b=c\)
Bài này có bạn giải rồi:
Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)
Điều này hiển nhiên đúng do:
\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)
\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)
Dấu "=" xảy ra khi \(a=b=c\)
Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)
\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)
Thật vậy, ta có:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)
Vậy ta cần chứng minh:
\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c
Áp dụng bất đẳng thức Schwarz và AM - GM ta có:
\(VT=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{3\left(a+b+c\right)}{\sqrt{3\left(a^2+b^2+c^2\right)}}-\dfrac{2\left(a+b+c\right)}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(\ge2\sqrt{\dfrac{3\left(a+b+c\right)^3}{\left(ab+bc+ca\right)\sqrt{3\left(a^2+b^2+c^2\right)}}}-\dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(=2\sqrt[4]{\dfrac{3\left(a+b+c\right)^6}{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}}-2\)
\(\ge2\sqrt[4]{\dfrac{3\left(a+b+c\right)^6}{\dfrac{\left(ab+bc+ca+ab+bc+ca+a^2+b^2+c^2\right)^3}{27}}}-2\)
\(=6-2=4=VP\left(đpcm\right)\).
Đặt vế trái của biểu thức là P
\(P=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}+\dfrac{\left(a+b+c\right)^2}{6\left(ab+bc+ca\right)}+\dfrac{\left(a+b+c\right)^2}{6\left(ab+bc+ca\right)}+\dfrac{a+b+c}{\sqrt{12\left(a^2+b^2+c^2\right)}}+\dfrac{a+b+c}{\sqrt{12\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{432\left(ab+bc+ca\right)\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}}\)
\(P\ge2+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{432\left(\dfrac{2ab+2bc+2ca+a^2+b^2+c^2}{3}\right)^3}}\)
\(P\ge2+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{16\left(a+b+c\right)^6}}=4\)
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{b}+\frac{1}{c}\geq \frac{4}{b+c}\)
\(\Rightarrow \frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}(1)\)
Hoàn toàn tương tự: \(\frac{b}{c}+\frac{b}{a}\geq \frac{4b}{c+a}(2)\)
\(\frac{c}{a}+\frac{c}{b}\geq \frac{4c}{a+b}(3)\)
Lấy \((1)+(2)+(3)\Rightarrow \frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(\Leftrightarrow \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c$