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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 $a=b=c$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sum\left(\dfrac{a^2}{b}\right)=\sum\left(\dfrac{a^4}{a^2b}\right)\ge\dfrac{\sum^2a^2}{\sum a^2b}\ge\dfrac{\sum^2a^2}{\sqrt{\sum a^2\cdot\sum a^2b^2}}\)
\(\Rightarrow\sum\left(\dfrac{a^2}{b}\right)\ge\dfrac{\sum^2a^2}{\sqrt{\dfrac{1}{3}\sum a^2\cdot\sum^2a^2}}=\sqrt{3\sum a^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Dự đoán dấu "=" xảy ra khi \(a=b=c=1\) ta tính được giá trị là \(9\)
Ta sẽ chứng minh nó là GTLN
Thật vậy ta cần chứng minh
\(\Sigma\dfrac{11a+4b}{4a^2-ab+2b^2}\le\dfrac{3\left(ab+ac+bc\right)}{abc}\)
\(\LeftrightarrowΣ\left(\dfrac{3}{a}-\dfrac{11a+4b}{4a^2-ab+2b^2}\right)\ge0\)
\(\LeftrightarrowΣ\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}\ge0\)
\(\LeftrightarrowΣ\left(\dfrac{\left(a-b\right)\left(a-6b\right)}{a\left(4a^2-ab+2b^2\right)}+\dfrac{1}{b}-\dfrac{1}{a}\right)\ge0\)
\(\LeftrightarrowΣ\dfrac{\left(a-b\right)^2\left(a+b\right)}{ab\left(4a^2-ab+2b^2\right)}\ge0\) (luôn đúng)
Bài 2:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(a^5+b^2+c^2\right)\left(\dfrac{1}{a}+b^2+c^2\right)\ge\left(a^2+b^2+c^2\right)^2\)
\(\Rightarrow\dfrac{1}{a^5+b^2+c^2}\le\dfrac{\dfrac{1}{a}+b^2+c^2}{\left(a^2+b^2+c^2\right)^2}\)
Tương tự rồi cộng theo vế ta có:
\(Σ\dfrac{1}{a^5+b^2+c^2}\le\dfrac{Σ\dfrac{1}{a}+2Σa^2}{\left(a^2+b^2+c^2\right)^2}\)
Ta chứng minh \(Σ\dfrac{1}{a}+2\left(a^2+b^2+c^2\right)\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\) - BĐT cuối đúng
Vậy ta có ĐPCM. Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 3:
Từ \(a+b+c=3abc\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=3\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow xy+yz+xz=3\) và BĐT cần chứng minh là
\(x^3+y^3+z^3\ge3\). Áp dụng BĐT AM-GM ta có:
\(x^3+x^3+1\ge3\sqrt[3]{x^3\cdot x^3\cdot1}=3x^2\)
Tương tự có: \(y^3+y^3+1\ge3y^2;z^3+z^3+1\ge3z^2\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(x^3+y^3+z^3\right)+3\ge3\left(x^2+y^2+z^2\right)\)
Lại có BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge3\left(xy+yz+xz\right)=9\left(xy+yz+xz=3\right)\)
\(\Rightarrow2\left(x^3+y^3+z^3\right)+3\ge9\Rightarrow2\left(x^3+y^3+z^3\right)\ge6\)
\(\Rightarrow x^3+y^3+z^3\ge3\). BĐT cuối đúng nên ta có ĐPCM
Đẳng thức xảy ra khi \(a=b=c=1\)
T/b:Vâng, rất giỏi
![](https://rs.olm.vn/images/avt/0.png?1311)
Hân đz đã đến :v giờ lm nha
Ta có: \(a^3=a\cdot a^2\)
\(\Rightarrow a^3+a\cdot b^2=a\cdot a^2+a\cdot b^2=a\left(a^2+b^2\right)\)
\(\Rightarrow\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\)(*)
Ta có: \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(\Rightarrow\dfrac{ab^2}{a^2+b^2}\le\dfrac{ab^2}{2ab}=\dfrac{b}{2}\)
\(\Rightarrow\dfrac{a^3}{a^2+b^2}\ge a-\dfrac{b}{2}\)
Tương tự: \(\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\); \(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)
Cộng 3 bđt trên ta có:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge a+b+c-\dfrac{b}{2}-\dfrac{c}{2}-\dfrac{a}{2}=\dfrac{a+b+c}{2}\)
''='' xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Cô-si:
\(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2}{b+c}\cdot\dfrac{\left(b+c\right)}{4}}=2\sqrt{\dfrac{a^2}{4}}=a\)
Chứng minh tương tự :
\(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b;\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{1}{4}\left(2a+2b+2c\right)\ge a+b+c\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge a+b+c-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{2}\) Dấu= xảy ra \(\Leftrightarrow a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{a^4}{a\left(b+c\right)}+\dfrac{b^4}{b\left(a+c\right)}+\dfrac{c^4}{c\left(a+b\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2ab+2ac+2bc}\)
\(A\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+a^2+c^2+b^2+c^2}=\dfrac{a^2+b^2+c^2}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Cô-si vào các số dương a,b,c:
\(\dfrac{a^2}{b}+b\ge2\sqrt{\dfrac{a^2}{b}\cdot b}=2\sqrt{a^2}=2a\Rightarrow\dfrac{a^2}{b}\ge2a-b\)
Chứng minh tương tự ta được:
\(\dfrac{b^2}{c}\ge2b-c;\dfrac{c^2}{a}\ge2c-a\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge2a+2b+2c-a-b-c=a+b+c\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Ta có: \(a+b+c=3\)
Áp dụng BĐT Cauchy - Schwarz ta có:
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{b+c+c+a+a+b}\)
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\cdot\left(a+b+c\right)}\)
\(P=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{3^2}{2\cdot3}=\dfrac{3}{2}\)
__________________
Nhắc lại BĐT Cauchy - Schwarz:
\(\dfrac{x^2_1}{a_1}+\dfrac{x^2_2}{a_2}+\dfrac{x^2_3}{a_3}+...+\dfrac{x^2_n}{a_n}\ge\dfrac{\left(x_1+x_2+...+x_n\right)^2}{a_1+a_2+...+a_n}\)
(p/s: bạn xem lại để nhé !)