1) Áp dụng bđt Cauchy:

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge2\sqrt{\dfrac{1}{a^2b^2}}=\dfrac{2}{ab}\)

Xong

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)

Tương tự:

\(\frac{1}{b^3(c+a)}+\frac{b(c+a)}{4}\geq \frac{1}{b}=ac\)

\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq \frac{1}{c}=ab\)

Cộng theo vế:

\(\Rightarrow \text{VT}+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{ab+bc+ac}{2}\)

Tiếp tục áp dụng AM-GM: \(ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}=3\)

\(\Rightarrow \text{VT}\ge \frac{3}{2}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c=1$

Lời giải:

Đặt vế trái là $A$

Áp dụng BĐT Bunhiacopxky:

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

\(\Leftrightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{36}{a+2b+3c}\)

Hoàn toàn TT:

\(\frac{1}{b}+\frac{2}{c}+\frac{3}{a}\geq \frac{36}{b+2c+3a}\)

\(\frac{1}{c}+\frac{2}{a}+\frac{3}{b}\geq \frac{36}{c+2a+3b}\)

Cộng theo vế:

\(\Rightarrow 6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 36A\)

\(\Rightarrow A\leq \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Theo đkđb: \(ab+bc+ac=abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Do đó: \(A\leq \frac{1}{6}< \frac{3}{16}\) (đpcm)

đặt ab=x, bc=y, ac=z

suy ra \(x^3+y^3+z^3=3xyz\)

pt thanh nhân tử \(\left(x+y+z\right)\left(x^2+y^2+z^2-xz-xy-yz\right)=0\)

do x,y,z>0suy ra x+y+z>0

nên suy ra \(x^2+y^2+z^2-xz-yz-xy=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xz-2xy-2yz=0\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

suy ra x=y=z

thế vào pt ta có dpcm

Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.

mk viết nhầm

\(ab+bc+ca=1\)

bn giúp mk với

Lời giải:

Áp dụng BĐT Bunhiacopxky với các số thực \(a,b,c\)

\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\right)\left ( \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right )\geq \left ( \frac{a}{c}+\frac{b}{a}+\frac{c}{b} \right )^2\)

\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}\right)\left ( \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right )\geq \left ( \frac{b}{c}+\frac{c}{a}+\frac{a}{b} \right )^2\)

Cộng hai vế trên thu được:

\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\right)\left(\frac{a+c}{b}+\frac{b+c}{a}+\frac{a+b}{c}\right)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)

Tiếp tục áp dụng Bunhiacopxky:

\([\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2](1+1)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)

Suy ra:

\(\left(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\right)\left(\frac{a+c}{b}+\frac{b+c}{a}+\frac{a+b}{c}\right)\geq \frac{1}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)^2\)

\(\Leftrightarrow \frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\geq \frac{1}{2}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)