\(a+b+ab=3\Leftrightarrow a+b+\frac{1}{4}\left(a+b\right)^2\ge3\)

\(\Leftrightarrow\left(a+b\right)^2+4\left(a+b\right)-12\ge0\)

\(\Leftrightarrow\left(a+b+6\right)\left(a+b-2\right)\ge0\)

\(\Leftrightarrow a+b\ge2\)

a/ \(\frac{a^4}{b}+\frac{b^4}{a}\ge\frac{\left(a^2+b^2\right)^2}{a+b}\ge\frac{\left(a+b\right)^4}{4\left(a+b\right)}=\frac{\left(a+b\right)^3}{4}\ge\frac{2^3}{4}=2\)

Dấu "=" xảy ra khi \(a=b=1\)

b/ \(\frac{a^4}{b^2}+\frac{b^4}{a^2}\ge\frac{\left(a^2+b^2\right)^2}{b^2+a^2}=a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\ge\frac{1}{2}.2^2=2\)

Dấu "=" xảy ra khi \(a=b=1\)

Bạn xem lại đề bài, cả 2 câu đều sai

Cho \(a=b=0,5\) ; \(c=2\) sau đó thay a;b vào 2 BĐT đều sai hết

Muốn chứng minh được 2 BĐT này thì điều kiện phải là \(a+b=2\) ; \(a;b>0\) ko liên quan gì tới c ở đây hết

\(2=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow a+b\le2\)

\(\left(a^3+b^3\right)\left(a+b\right)\ge\left(a^2+b^2\right)^2=4\)

\(a^3+b^3\ge\dfrac{4}{a+b}\ge\dfrac{4}{2}=2\)

\(a^3+a\ge2a^2\);\(b^3+b\ge2b^2\)

\(a^3+b^3\ge2\left(a^2+b^2\right)-\left(a+b\right)\ge4-\sqrt{2\left(a^2+b^2\right)}=4-2=2\)

các bạn hãy áp dụng bất đẳng thức Cô-si

Có \(a+\frac{1}{b\left(a-b\right)^2}=\left(a-b\right)+b+\frac{1}{b\left(a-b\right)^2}=\frac{a-b}{2}+\frac{a-b}{2}+b+\frac{1}{b\left(a-b\right)^2}\)

Áp dụng BĐT Cosi cho 4 số ta có:

\(\frac{a-b}{2}+\frac{a-b}{2}+b+\frac{1}{b\left(a-b\right)^2}\ge4\sqrt[4]{\frac{a-b}{2}\cdot\frac{a-b}{2}\cdot b\cdot\frac{1}{b\left(a-b\right)^2}}\)

\(=4\cdot\sqrt[4]{\frac{1}{4}}=1\cdot\frac{\sqrt{1}}{2}=2\sqrt{2}\)

\(\Rightarrow a+\frac{1}{b\left(a-b\right)^2}\ge2\sqrt{2}\)

Dấu "=" xảy ra khi \(\frac{a-b}{2}=b\)

\(\Leftrightarrow\frac{a}{2}=\frac{3b}{2}\Leftrightarrow a=3b\)

Cách giải: Linh Vy. Trình bày: Nhật Quỳnh

BĐT cần chứng minh tương đương

\(\dfrac{3a^2+2ab+3b^2}{a+b}-2\left(a+b\right)\ge2\sqrt{2\left(a^2+b^2\right)}-2\left(a+b\right)\)

\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{a+b}\ge\dfrac{8\left(a^2+b^2\right)-4\left(a+b\right)^2}{2\sqrt{2\left(a^2+b^2\right)}+2\left(a+b\right)}\)

\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{a+b}\ge\dfrac{2\left(a-b\right)^2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)

\(\Leftrightarrow\left(a-b\right)^2\left(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\right)\ge0\)

ta phải chứng minh

\(\dfrac{1}{a+b}-\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\ge0\)

\(\Leftrightarrow\dfrac{1}{a+b}\ge\dfrac{2}{\sqrt{2\left(a^2+b^2\right)}+a+b}\)

\(\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}+a+b\ge2\left(a+b\right)\Leftrightarrow\sqrt{2\left(a^2+b^2\right)}\ge a+b\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)

=> đpcm

Áp dụng bđt AM - GM ta có :

\(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{a^2}}\ge\sqrt{2\frac{a^2}{b^2}}+\sqrt{2\frac{b^2}{a^2}}=\sqrt{2}\frac{a}{b}+\sqrt{2}\frac{b}{a}\)

\(=\sqrt{2}\left(\frac{a}{b}+\frac{b}{a}\right)\ge\sqrt{2}.2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\sqrt{2}\)

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge2\)

\(\Leftrightarrow a^2+2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+2\ge2\)

<=> Sai đề

Đặt \(A=\frac{ab}{\left(b-c\right)\left(c-a\right)}+\frac{bc}{\left(c-a\right)\left(a-b\right)}+\frac{ca}{\left(b-c\right)\left(a-b\right)}=-1\)

\(\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)^2\ge0\)

\(\Leftrightarrow\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2+2A\ge0\)

\(\Leftrightarrow\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\ge2\)

\(\ge2\)

chúc bn học tốt nhớ tích và kb với mk nha ^^