a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

\(\left(a+1\right)\left(b+1\right)=4ab\Leftrightarrow\left(\dfrac{1}{a}+1\right)\left(\dfrac{1}{b}+1\right)=4\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow\left(x+1\right)\left(y+1\right)=4\Rightarrow xy=3-x-y\)

\(P=\dfrac{x}{\sqrt{x^2+3}}+\dfrac{y}{\sqrt{y^2+3}}\le\dfrac{x}{\sqrt{\dfrac{\left(x+3\right)^2}{4}}}+\dfrac{y}{\sqrt{\dfrac{\left(y+3\right)^2}{4}}}=\dfrac{2x}{x+3}+\dfrac{2y}{y+3}\)

\(P\le\dfrac{4xy+6x+6y}{\left(x+3\right)\left(y+3\right)}=\dfrac{4xy+6x+6y}{xy+3x+3y+9}=\dfrac{4\left(3-x-y\right)+6x+6y}{3-x-y+3x+3y+9}=\dfrac{2x+2y+12}{2x+2y+12}=1\)

\(P_{max}=1\) khi \(x=y=1\) hay \(a=b=1\)

Hi vọng là tìm GTLN:

Không mất tính tổng quát, giả sử b, c cùng phía với 1 \(\Rightarrow\left(b-1\right)\left(c-1\right)\ge0\Leftrightarrow bc\ge b+c-1\).

Áp dụng bất đẳng thức AM - GM ta có: 

\(4=a^2+b^2+c^2+abc\ge a^2+2bc+abc\Leftrightarrow2bc+abc\le4-a^2\Leftrightarrow bc\left(a+2\right)\le\left(2-a\right)\left(a+2\right)\Leftrightarrow bc+a\le2\)

\(\Rightarrow a+b+c\le3\).

Áp dụng bất đẳng thức Schwarz ta có:

\(P\le\dfrac{ab}{9}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)+\dfrac{bc}{9}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)+\dfrac{ca}{9}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)=\dfrac{1}{9}.3\left(a+b+c\right)=\dfrac{1}{3}\left(a+b+c\right)\le1\).

Đẳng thức xảy ra khi a = b = c = 1.

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+5=11\)

Vậy: P(min)=11  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

-Đề thiếu?

Lời giải:

Áp dụng BĐT AM-GM:
\(P=\sum \sqrt{\frac{ab}{c+ab}}=\sum \sqrt{\frac{ab}{c(a+b+c)+ab}}=\sum \sqrt{\frac{ab}{(c+a)(c+b)}}\)

\(\leq \sum \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)

Vậy $P_{\max}=\frac{3}{2}$ khi $a=b=c=\frac{1}{3}$

Em tham khảo ở đây:

xét các số thực a,b,c (a≠0) sao cho phương trình ax2+bx+c=0 có 2 nghiệm m, n thỏa mãn \(0\le m\le1;0\le m\le1\). tìm GTN... - Hoc24

Max thì đơn giản thôi em:

Do \(0\le m;n\le1\Rightarrow0< 2-mn\le2\)

\(\Rightarrow M=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{m+n+1}=2\)

\(M_{max}=2\) khi \(mn=0\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Rightarrow2\ge3x^2+2y^2+2z^2+y^2+z^2\) 

\(\Leftrightarrow2\ge3\left(x^2+y^2+z^2\right)\)

Có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\le2\)

\(\Rightarrow\)\(A^2\le2\) \(\Leftrightarrow A\in\left[-\sqrt{2};\sqrt{2}\right]\)

minA=-1\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=-\sqrt{2}\\x=y=z\end{matrix}\right.\)  \(\Rightarrow x=y=z=-\dfrac{\sqrt{2}}{3}\)

maxA=1\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=\sqrt{2}\\x=y=z\end{matrix}\right.\) \(\Rightarrow x=y=z=\dfrac{\sqrt{2}}{3}\)