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\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)\ge\dfrac{9}{2}\)
\(\Rightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)
\(\Rightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge9\)
\(\Rightarrow\left(a+b+c+a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge9\)
Đặt: \(\left\{{}\begin{matrix}a+b=x\\b+c=y\\c+a=z\end{matrix}\right.\) Khi đó bất đẳng thức trở thành:
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\) (đúng theo AM-GM)
Vậy bất đẳng thức cần chứng minh đúng
Dấu "=" xảy ra khi: \(a=b=c>0\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Ta có \(\dfrac{a^2}{a+b^2}=a-\dfrac{ab^2}{a+b^2}\ge a-\dfrac{ab^2}{2b\sqrt{a}}=a-\dfrac{ab}{2\sqrt{a}}\)
Thiết lập tương tự và thu lại ta có :
\(VT\ge3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\)
Xét \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}=\sqrt{\dfrac{a^2b^2}{4a}}+\sqrt{\dfrac{b^2c^2}{4b}}+\sqrt{\dfrac{a^2c^2}{4c}}\)
Áp dụng bđt Cauchy ta có \(\sqrt{\dfrac{a^2b^2}{4a}}=\sqrt{\dfrac{ab}{2a}.\dfrac{ab}{2}}\le\dfrac{\dfrac{b}{2}+\dfrac{ab}{2}}{2}\)
Thiết lập tương tự và thu lại ta có :
\(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{\dfrac{a+b+c}{2}+\dfrac{ab+bc+ac}{2}}{2}=\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\left(1\right)\)
Theo hệ quả của bđt Cauchy ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\le\dfrac{\dfrac{3}{2}+\dfrac{3}{2}}{2}=\dfrac{3}{2}\left(2\right)\)
Từ ( 1 ) và ( 2 ) ta có \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{3}{2}\)
\(\Rightarrow3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=1\)
3.
\(\dfrac{2a^2}{b^2}+2\dfrac{b^2}{c^2}+2\dfrac{c^2}{a^2}\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
áp dụng bất đẳng thức cosi
+ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\dfrac{a}{c}\)
......
tương tự với 2 cái sau
Chứng minh : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(x^2-y^2-xy-xz+2yz\right)^2+\left(y^2-z^2-yz-xy+2xz\right)^2+\left(z^2-x^2-xz-yz+2xy\right)^2\right)\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge a-\dfrac{a^2b}{2\sqrt{ab}}=a-\dfrac{\sqrt{a^3b}}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b}{bc+1}\ge b-\dfrac{\sqrt{b^3c}}{2};\dfrac{c}{ca+1}\ge c-\dfrac{\sqrt{c^3a}}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge3-\dfrac{1}{2}\left(\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
\(\sum\dfrac{a^3}{a^2+b^2}=a+b+c-\dfrac{ab^2}{a^2+b^2}-\dfrac{bc^2}{b^2+c^2}-\dfrac{ca^2}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\) Dấu "=" xảy ra khi: \(a=b=c\)