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Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)
\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Hay 1 cách khác :AM-GM
\(\dfrac{b}{a^2}+\dfrac{c}{a^2}+\dfrac{1}{b}+\dfrac{1}{c}\ge4\sqrt[4]{\dfrac{1}{a^4}}=\dfrac{4}{a}\)
Tương tự là ta có ngay đpcm
Một cách đơn giản nhất tương đương ( hay còn gọi là SOS)
\(BĐT\Leftrightarrow\sum\dfrac{b+c-2a}{a^2}\ge0\)
\(\Leftrightarrow\sum\left(\dfrac{b-a}{a^2}+\dfrac{c-a}{a^2}\right)\ge0\)
Nhóm lại: \(\Leftrightarrow\sum\left(\dfrac{a-b}{b^2}+\dfrac{b-a}{a^2}\right)\ge0\)
\(\Leftrightarrow\sum\left(a-b\right)^2.\left(\dfrac{a+b}{a^2b^2}\right)\ge0\)(đúng)
Vậy BĐT được chứng minh.
Dấu = xảy ra khi a=b=c
a) Sai với \(a=1,b=2\)
b)
Thực hiện biến đổi tương đương:
\(\frac{a}{3b}+\frac{b(a+b)}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}+\frac{b(a+b)+a^2}{a^2+ab+b^2}-\frac{a^2}{a^2+ab+b^2}\geq 1\)
\(\Leftrightarrow \frac{a}{3b}-\frac{a^2}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{1}{3b}-\frac{a}{a^2+ab+b^2}\geq 0\)
\(\Leftrightarrow \frac{a^2+ab+b^2-3ab}{3b(a^2+ab+b^2)}\geq 0\)
\(\Leftrightarrow \frac{(a-b)^2}{3b(a^2+ab+b^2)}\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b$
c) BĐT sai với \(a=1,b=2\)
ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có
\(\Sigma\dfrac{1}{\left(a+b\right)c}\le\Sigma\dfrac{1}{2\sqrt{c}}=\dfrac{\Sigma\sqrt{ab}}{2}\le\dfrac{\Sigma a}{2}\)(đpcm)
Hình như thế này mới đúng chứ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Áp dụng BĐT Cosi:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2.\dfrac{a}{c};\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2.\dfrac{b}{a};\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2.\dfrac{c}{b}\)
\(\Rightarrow2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)
Đẳng thức xảy ra khi \(a=b=c>0\)
\(VT=\dfrac{a}{b\left(b^2+a\right)}+\dfrac{b}{c\left(c^2+b\right)}+\dfrac{c}{a\left(a^2+c\right)}\)
\(VT=\dfrac{a+b^2-b^2}{b\left(b^2+a\right)}+\dfrac{b+c^2-c^2}{c\left(c^2+b\right)}+\dfrac{c+a^2-a^2}{a\left(a^2+c\right)}\)
\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)
\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{b}{b^2+a}+\dfrac{c}{c^2+b}+\dfrac{a}{a^2+c}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{b}{b^2+a}\le\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)
Thiết lập tương tự và thu lại tao có
\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\sqrt{\dfrac{1}{a}}\le\dfrac{\dfrac{1}{a}+1}{2}\)
Tương tự ta có
\(\sqrt{\dfrac{1}{b}}\le\dfrac{\dfrac{1}{b}+1}{2};\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{c}+1}{2}\)
Thu lại ta có
\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\right)\)
\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}-\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Ta có đánh giá sau với a không âm:
\(\dfrac{a}{1+a^2}\le\dfrac{36a+3}{50}\)
Thật vậy, BĐT tương đương:
\(\left(36a+3\right)\left(a^2+1\right)\ge50a\)
\(\Leftrightarrow\left(3a-1\right)^2\left(4a+3\right)\ge0\) (luôn đúng)
Tương tự: \(\dfrac{b}{1+b^2}\le\dfrac{36b+3}{50}\) ; \(\dfrac{c}{1+c^2}\le\dfrac{36c+3}{50}\)
Cộng vế: \(VT\le\dfrac{36\left(a+b+c\right)+9}{50}=\dfrac{9}{10}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)
Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)
\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)
\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)
Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)
Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(A=\dfrac{a}{b+c}+\dfrac{b+c}{a}+\dfrac{b}{c+a}+\dfrac{c+a}{b}+\dfrac{c}{a+b}+\dfrac{a+b}{c}\)
\(A=\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right)\)
\(A\ge\dfrac{3}{2}+\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\) (bất đẳng thức Nesbit)
\(A\ge\dfrac{3}{2}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}\)
\(A\ge\dfrac{3}{2}+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng bất đẳng thức AM-GM cho 2 số dương ta có:
\(A\ge\dfrac{3}{2}+2\sqrt{\dfrac{ab}{ab}}+2\sqrt{\dfrac{ac}{ac}}+2\sqrt{\dfrac{bc}{bc}}\)
\(A\ge\dfrac{3}{2}+2+2+2=\dfrac{15}{2}\left(đpcm\right)\)
Dấu"=" xảy ra khi: \(a=b=c\)