1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\cdot VT\ge\left(1+1+1\right)^2\)

Lại có BĐT \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\)

\(\Rightarrow\sqrt{a^2+b^2}\ge\sqrt{\dfrac{\left(a+b\right)^2}{2}}=\dfrac{a+b}{\sqrt{2}}\)

Tương tự cũng có: \(\sqrt{b^2+c^2}\ge\dfrac{b+c}{\sqrt{2}};\sqrt{c^2+a^2}\ge\dfrac{c+a}{\sqrt{2}}\)

\(\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\dfrac{2\left(a+b+c\right)}{\sqrt{2}}\)

\(\ge\dfrac{2\cdot\sqrt{3\left(ab+bc+ca\right)}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}\)

\(\Rightarrow VT\ge\dfrac{\left(1+1+1\right)^2}{\dfrac{2\sqrt{3}}{\sqrt{2}}}=\dfrac{3\sqrt{6}}{2}\)

Xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Bữa trước ko để ý a,b,c ko âm với ngược dấu sai thê thảm =))

Dự đoán \(a=b=1\) và \(c=0\) thì tính được \(2+\frac{1}{\sqrt2}\)

Ta sẽ chứng minh nó là GTNN.Thật vậy cần chứng minh

\(\sqrt{\dfrac{ab+bc+ca}{a^2+b^2}}+\sqrt{\dfrac{ab+bc+ca}{b^2+c^2}}+\sqrt{\dfrac{ab+bc+ca}{a^2+c^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

Khôn mất tính tổng quá giả sử \(c=\min\{a,b,c\}\). Khi đó:

\(\dfrac{ab+ac+bc}{a^2+b^2}-\dfrac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\dfrac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\ge0\)

Tương tự cũng có:

\(\dfrac{ab+ac+bc}{a^2+c^2}-\dfrac{b+c}{a+c}=\dfrac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\ge0\)

Và \(\dfrac{ab+ac+bc}{b^2+c^2}-\dfrac{a+c}{b+c}=\dfrac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\ge0\)

Đặt \(\dfrac{a+c}{b+c}=x^2;\dfrac{b+c}{a+c}=y^2\left(x,y>0\right)\)\(\Rightarrow xy=1\) và ta có:

\(x+y+\dfrac{1}{\sqrt{x^2+y^2}}\ge2+\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x+y-2\sqrt{xy}\ge\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{x^2+y^2}}\)

\(\Leftrightarrow(\sqrt{x}-\sqrt{y})^2\ge\dfrac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})}\)

\(\Leftrightarrow\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt{2})\ge(\sqrt{x}+\sqrt{y})^2\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\ge x+y\)

\(=\dfrac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\ge\dfrac{1}{2}(\sqrt{x}+\sqrt{y})^2\)

Vậy cần chứng minh \(\sqrt{x^2+y^2}+\sqrt{2}\ge2\)

Đúng theo AM-GM:\(\sqrt{x^2+y^2}+\sqrt{2}\ge\sqrt{2xy}+\sqrt{2}=2\sqrt{2}>2\)

mong mn giúp mk vs

Bạn tham khảo:

Cho \(a,b,c>\dfrac{25}{4}.\)Tìm GTNN của \(Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\) - Hoc24

cái kia là \(3\sqrt{\dfrac{1}{a}+\dfrac{9}{b}+\dfrac{25}{c}}\)

\(\left(a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}\right)\left(1+3+5\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow3\sqrt{a^2+\dfrac{b^2}{3}+\dfrac{c^2}{5}}\ge a+b+c\)

\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{1}{a}+\dfrac{3^2}{b}+\dfrac{5^2}{c}}\)

\(\Rightarrow P\ge\dfrac{2}{3}\left(a+b+c\right)+3\sqrt{\dfrac{\left(1+3+5\right)^2}{a+b+c}}=\dfrac{2}{3}\left(a+b+c\right)+\dfrac{27}{\sqrt{a+b+c}}\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(a+b+c\right)+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{27}{2\sqrt{a+b+c}}+\dfrac{1}{6}\left(a+b+c\right)\)

\(\Rightarrow P\ge3\sqrt[3]{\dfrac{27^2\left(a+b+c\right)}{2^3\left(a+b+c\right)}}+\dfrac{1}{6}.9=15\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;3;5\right)\)

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

mong mn giúp mk vs