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\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)
xong AM -GM
Đang học Bunyakovsky đúng hong :D
1)
\(S=\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ac+a^2}\)
\(S^2=\left(\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ac+a^2}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(a^2+4ab+b^2+b^2+4bc+c^2+c^2+4ac+a^2\right)\)
\(=3.2\left(a^2+b^2+c^2+2ab+2bc+2ac\right)=6.\left(a+b+c\right)^2=6.6^2=216\)
\(\Leftrightarrow S\le6\sqrt{6}."="\Leftrightarrow a=b=c=2\)
2) \(M^2=\left(\sqrt{x+1}+\sqrt{y+1}\right)^2\le\left(1^2+1^2\right)\left(x+1+y+1\right)=2.8=16\)
\(M\le4."="\Leftrightarrow x=y=3\)
3)
\(S=ab+2\left(a+b\right)\le\dfrac{\left(a+b\right)^2}{4}+\dfrac{8\left(a+b\right)}{4}\)
\(=\dfrac{\left(a+b\right)^2+8\left(a+b\right)}{4}\)
\(\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)=2\Leftrightarrow a+b\le\sqrt{2}\)
\(\dfrac{\left(a+b\right)^2+8\left(a+b\right)}{4}\le\dfrac{2+8\sqrt{2}}{4}=\dfrac{1+4\sqrt{2}}{2}\)
\(S\le\dfrac{1+4\sqrt{2}}{2}."="\Leftrightarrow a=b=\dfrac{1}{\sqrt{2}}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{\sqrt{3a^2+4ab+b^2}}=\dfrac{1}{\sqrt{\left(a+b\right)\left(3a+b\right)}}=\dfrac{\sqrt{2}}{\sqrt{\left(2a+2b\right)\left(3a+b\right)}}\)
\(\ge\dfrac{\sqrt{2}}{\dfrac{2a+2b+3a+b}{2}}=\dfrac{\sqrt{2}}{\dfrac{5a+3b}{2}}=\dfrac{2\sqrt{2}}{5a+3b}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{1}{\sqrt{3b^2+4bc+c^2}}\ge\dfrac{2\sqrt{2}}{5b+3c};\dfrac{1}{\sqrt{3c^2+4ca+a^2}}\ge\dfrac{2\sqrt{2}}{5c+3a}\)
Cộng theo vế 3 BĐT trên ta có:
\(P\ge\dfrac{2\sqrt{2}}{5a+3b}+\dfrac{2\sqrt{2}}{5b+3c}+\dfrac{2\sqrt{2}}{5c+3a}\)
\(\ge\dfrac{18\sqrt{2}}{8\left(a+b+c\right)}=\dfrac{18\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}\)
Xảy ra khi \(a=b=c=\dfrac{1}{3}\)
mình đánh nhầm, đề là cho a,b,c là các số thực dương tổng bằng 1
Có \(\sqrt{a^2+4ab+b^2}=\sqrt{\left(\frac{3}{2}a^2+3ab+\frac{3}{2}b^2\right)-\left(\frac{1}{2}a^2-ab+\frac{1}{2}b^2\right)}\)
\(=\sqrt{\frac{3}{2}\left(a+b\right)^2-\frac{1}{2}\left(a-b\right)^2}\le\sqrt{\frac{3}{2}\left(a+b\right)^2}=\sqrt{\frac{3}{2}}\left(a+b\right)\)
Tương tự, ta có : \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}\left(b+c\right);\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}\left(c+a\right)\)
\(\Rightarrow\)\(S\le\sqrt{\frac{3}{2}}\left(a+b\right)+\sqrt{\frac{3}{2}}\left(b+c\right)+\sqrt{\frac{3}{2}}\left(c+a\right)=\sqrt{\frac{3}{2}}.2\left(a+b+c\right)=6\sqrt{6}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)
a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)
\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)
\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)
\(a^2+ab+b^2=\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{4}\left(a+b\right)^2=\dfrac{3}{4}\left(a+b\right)^2\)
Tương tự, ta có:
\(M\ge\dfrac{\sqrt{3}}{2}\left(a+b\right)+\dfrac{\sqrt{3}}{2}\left(b+c\right)+\dfrac{\sqrt{3}}{2}\left(c+a\right)=\sqrt{3}\left(a+b+c\right)=3\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Vì \(a;b;c>0\Rightarrow2ab\le\frac{\left(a+b\right)^2}{2}\) thay vào \(\sqrt{a^2+4ab+b^2}\)ta có:
\(\sqrt{a^2+4ab+b^2}=\sqrt{\left(a+b\right)^2+2ab}\)
\(\le\sqrt{\left(a+b\right)^2+\frac{\left(a+b\right)^2}{2}}=\sqrt{\frac{3\left(a+b\right)^2}{2}}=\left(a+b\right).\sqrt{\frac{3}{2}}\)
Tương tự: \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}.\left(b+c\right)\)
\(\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}.\left(c+a\right)\)
\(\Rightarrow P\le\sqrt{\frac{3}{2}}.\left(a+b\right)+\sqrt{\frac{3}{2}}.\left(b+c\right)+\sqrt{\frac{3}{2}}.\left(c+a\right)\)
\(\le\sqrt{\frac{3}{2}}.\left(2a+2b+2c\right)=\sqrt{\frac{3}{2}}.6=\sqrt{216}=6\sqrt{6}\)Vì a+b+c=6
Dấu = xảy ra khi a=b=c=2
Vây ......