VT= $\sqrt{(\frac{a}{\sqrt{2}}-b)^{2}+(\frac{a}{\sqrt{2}})^{2}}$ + $\sqrt{(b-\frac{\sqrt{3}}{2}c)^{2}+(\frac{c}{2})^{2}}$Đặt $\overrightarrow{u}$($\frac{a}{\sqrt{2}}$-b;$\frac{a}{\sqrt{2}}$) ; $\overrightarrow{v}$(b-$\frac{\sqrt{3}}{2}$c;$\frac{c}{2}$)VT=$\left| {\overrightarrow{u}} \right|$+$\left| {\overrightarrow{v}} \right|$$\geq$$\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$(1)Mà $\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$=$\sqrt{(\frac{a}{\sqrt{2}}-\frac{\sqrt{3}}{2}c)^{2}+(\frac{a}{\sqrt{2}}+\frac{c}{2})^{2}}$=$\sqrt{\frac{a^{2}}{2}+\frac{3}{4}c^{2}+\frac{a^{2}}{2}+\frac{c^{2}}{4}-\frac{\sqrt{3}}{\sqrt{2}}ac+\frac{1}{\sqrt{2}}ac}$=$\sqrt{a^{2}-\sqrt{2-\sqrt{3}}ac+c^{2}}$(2)Từ(1)&(2)$\Rightarrow$đpcm

VT = $ \sqrt{(\frac{a}{\sqrt{2}}-b)^{2}+(\frac{a}{\sqrt{2}})^{2}} $+$ \sqrt{(b-\frac{\sqrt{3}}{2}c)^{2}+(\frac{c}{2})^{2}}$Đặt$\overrightarrow{u}$($\frac{a}{\sqrt{2}}$-b;$\frac{a}{\sqrt{2}}$) ;$\overrightarrow{v}$(b-$\frac{\sqrt{3}}{2}$c;$\frac{c}{2}$)VT=$\left| {\overrightarrow{u}} \right|$+$\left| {\overrightarrow{v}} \right| $$\geq$$ \left| {\overrightarrow{u}+\overrightarrow{v}} \right|$(1)Mà$\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$=$\sqrt{(\frac{a}{\sqrt{2}}-\frac{\sqrt{3}}{2}c)^{2}+(\frac{a}{\sqrt{2}}+\frac{c}{2})^{2}}$=$\sqrt{\frac{a^{2}}{2}+\frac{3}{4}c^{2}+\frac{a^{2}}{2}+\frac{c^{2}}{4}-\frac{\sqrt{3}}{\sqrt{2}}ac+\frac{1}{\sqrt{2}}ac}$=$\sqrt{a^{2}-\sqrt{2-\sqrt{3}}ac+c^{2}}$(2)Từ(1)&(2)$\Rightarrow$đpcm

VT=$\sqrt{(\frac{a}{\sqrt{2}}-b)^{2}+(\frac{a}{\sqrt{2}})^{2}}$+$\sqrt{(b-\frac{\sqrt{3}}{2}c)^{2}+(\frac{c}{2})^{2}}$Đặt $\overrightarrow{u}$($\frac{a}{\sqrt{2}}$-b;$\frac{a}{\sqrt{2}}$) ; $\overrightarrow{v}$(b-$\frac{\sqrt{3}}{2}$c;$\frac{c}{2}$)VT=$\left| {\overrightarrow{u}} \right|$+$\left| {\overrightarrow{v}} \right|$$\geq$$\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$(1)Mà $\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$=$\sqrt{(\frac{a}{\sqrt{2}}-\frac{\sqrt{3}}{2}c)^{2}+(\frac{a}{\sqrt{2}}+\frac{c}{2})^{2}}$=$\sqrt{\frac{a^{2}}{2}+\frac{3}{4}c^{2}+\frac{a^{2}}{2}+\frac{c^{2}}{4}-\frac{\sqrt{3}}{\sqrt{2}}ac+\frac{1}{\sqrt{2}}ac}$=$\sqrt{a^{2}-\sqrt{2-\sqrt{3}}ac+c^{2}}$(2)Từ(1)&(2)$\Rightarrow$đpcm

VT= $\sqrt{(\frac{a}{\sqrt{2}}-b)^{2}+(\frac{a}{\sqrt{2}})^{2}}$ + $\sqrt{(b-\frac{\sqrt{3}}{2}c)^{2}+(\frac{c}{2})^{2}}$Đặt $\overrightarrow{u}$($\frac{a}{\sqrt{2}}$-b;$\frac{a}{\sqrt{2}}$) ; $\overrightarrow{v}$(b-$\frac{\sqrt{3}}{2}$c;$\frac{c}{2}$)VT=$\left| {\overrightarrow{u}} \right|$+$\left| {\overrightarrow{v}} \right|$$\geq$$\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$(1)Mà $\left| {\overrightarrow{u}+\overrightarrow{v}} \right|$=$\sqrt{(\frac{a}{\sqrt{2}}-\frac{\sqrt{3}}{2}c)^{2}+(\frac{a}{\sqrt{2}}+\frac{c}{2})^{2}}$=$\sqrt{\frac{a^{2}}{2}+\frac{3}{4}c^{2}+\frac{a^{2}}{2}+\frac{c^{2}}{4}-\frac{\sqrt{3}}{\sqrt{2}}ac+\frac{1}{\sqrt{2}}ac}$=$\sqrt{a^{2}-\sqrt{2-\sqrt{3}}ac+c^{2}}$(2)Từ(1)&(2)$\Rightarrow$đpcm

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$$\Rightarrow$abc$\leq$1P=$\Sigma$$\frac{a^{2}b}{1+a+b}$$\leq$$\Sigma$$\frac{a^{2}b}{3\sqrt[3]{ab}}$=$\Sigma$$\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc$\leq$1$\Rightarrow$$a^{2}$$b^{2}$$c^{2}$$\leq$1$\Rightarrow$P$\leq$$\frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$ $\Rightarrow$ abc $\leq$1P=$\Sigma$ $\frac{a^{2}b}{1+a+b}$ $\leq$ $\Sigma$ $\frac{a^{2}b}{3\sqrt[3]{ab}}$ = $\Sigma$ $\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc $\leq$ 1$\Rightarrow$ $a^{2}$$b^{2}$$c^{2}$ $\leq$1$\Rightarrow$P$\leq$$\frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1

Kẻ MN vuông góc vs CI tại N;MP vuông góc vs AB tại P(Tự vẽ hình)I$\epsilon$CI và CI vuông góc vs AB$\Rightarrow$CI:2x+y-10=0$\Rightarrow$MN:x-2y+5=0N=CI$\cap$MN$\Rightarrow$N(3;4)Mà MP:2x+y-$\frac{25}{2}$=0P=MP$\cap$AB$\Rightarrow$P(5;$\frac{5}{2}$)$\Rightarrow$S MNIP=MN.MP=$\frac{5}{2}$Gỉa sử:$\frac{BP}{BỊ}$=kTa có:$\frac{S MBP }{S CIB }$=$(\frac{BP}{BI})^{2}$=$k^{2} $$\frac{S CMN}{S CIB}$=$(\frac{BI-BP}{BI})^{2}$=$(1-k)^{2}$Ta có:S ABC=10$\Rightarrow$S CIB=5Mặt khác:S MNIP=S CIB- S MBP- S CMN$\Leftrightarrow$S CIB.$\left[ {1-(1-k)^{2}-k^{2}} \right]$=$\frac{5}{2}$$ \Leftrightarrow $k=$\frac{1}{2} $$\Rightarrow$$ \overrightarrow{BI}$=2$\overrightarrow{BP}$$\Rightarrow$B(6;3)$\Rightarrow$BC(BM):3x+4y-30=0C=BC$\cap$CI$\Rightarrow$C(2;6)I là tđ AB$\Rightarrow$A(2;1)

Kẻ MN vuông góc vs CI tại N;MP vuông góc vs AB tại P(Tự vẽ hình)I$\epsilon$CI và CI vuông góc vs AB$\Rightarrow$CI:2x+y-10=0$\Rightarrow$MN:x-2y+5=0N=CI$\cap$MN$\Rightarrow$N(3;4)Mà MP:2x+y-$\frac{25}{2}$=0P=MP$\cap$AB$\Rightarrow$P(5;$\frac{5}{2}$)$\Rightarrow$S MNIP=MN.MP=$\frac{5}{2}$Gỉa sử:$\frac{BP}{BI}$=kTa có:$\frac{S MBP }{S CIB }$=$(\frac{BP}{BI})^{2}$=$k^{2} $$\frac{S CMN}{S CIB}$=$(\frac{BI-BP}{BI})^{2}$=$(1-k)^{2}$Ta có:S ABC=10$\Rightarrow$S CIB=5Mặt khác:S MNIP=S CIB- S MBP- S CMN$\Leftrightarrow$S CIB.$\left[ {1-(1-k)^{2}-k^{2}} \right]$=$\frac{5}{2}$$ \Leftrightarrow $k=$\frac{1}{2} $$\Rightarrow$$ \overrightarrow{BI}$=2$\overrightarrow{BP}$$\Rightarrow$B(6;3)$\Rightarrow$BC(BM):3x+4y-30=0C=BC$\cap$CI$\Rightarrow$C(2;6)I là tđ AB$\Rightarrow$A(2;1)

Ta có:a$\sqrt{b^{3}+1}$+b$\sqrt{c^{3}+1}$+c$\sqrt{a^{3}+1}$=a$\sqrt{(b+1)(b^{2}-b+1)}$ + b$\sqrt{(c+1)(c^{2}-c+1)}$+c$\sqrt{(a+1)(a^{2}-a+1)}$$\leq$a.$\frac{b^{2}+2}{2}$+b.$\frac{c^{2}+2}{2}$+c.$\frac{a^{2}+2}{2}$=$\frac{ab^{2}+bc^{2}+ca^{2}}{2}$+3Ta phải cm:a$b^{2}$+b$c^{2}$+c$a^{2}$$\leq$4(1)Gỉa sử a$\leq$b$\leq$c,ta có:a(b-a)(b-c)$\leq$0$\Leftrightarrow$a$b^{2}$+c$a^{2}$$\leq$b$a^{2}$+abc$\Rightarrow$a$b^{2}$+b$c^{2}$+c$a^{2} $$\leq$b$a^{2}$+abc+b$c^{2}$=b($a^{2}$+ac+$c^{2}$)$\leq$b$(a+c)^{2}$=$\frac{1}{2}$.2b.$(3-b)^{2}$$ \leq$$\frac{1}{2}$.$(\frac{2b+3-b+3-b}{3})^{3}$=4$\Rightarrow$đpcmDấu''=''xra$\Leftrightarrow$a=0;b=1;c=2 và các hoán vị

Ta có:a$\sqrt{b^{3}+1}$+b$\sqrt{c^{3}+1}$+c$\sqrt{a^{3}+1}$=a$\sqrt{(b+1)(b^{2}-b+1)}$ + b$\sqrt{(c+1)(c^{2}-c+1)}$+c$\sqrt{(a+1)(a^{2}-a+1)}$$\leq$a.$\frac{b^{2}+2}{2}$+b.$\frac{c^{2}+2}{2}$+c.$\frac{a^{2}+2}{2}$=$\frac{ab^{2}+bc^{2}+ca^{2}}{2}$+3(1)Ta phải cm:a$b^{2}$+b$c^{2}$+c$a^{2}$$\leq$4Gỉa sử a$\leq$b$\leq$c,ta có:a(b-a)(b-c)$\leq$0(2)$\Leftrightarrow$a$b^{2}$+c$a^{2}$ $ \leq$b$a^{2}$+abc$\Leftrightarrow $a$b^{2}$+b$c^{2}$+c$a^{2} $$\leq$b$a^{2}$+abc+b$c^{2}$=b($a^{2}$+ac+$c^{2}$)$\leq$b$(a+c)^{2}$=$\frac{1}{2}$.2b.$(3-b)^{2}$$ \leq$$\frac{1}{2}$.$(\frac{2b+3-b+3-b}{3})^{3}$=4(3)$\Rightarrow$đpcmXét dấu''='' xra ở (1);(2);(3)$\Rightarrow$Dấu''=''xra$\Leftrightarrow$a=0;b=1;c=2 (và các hoán vị tùy theo cách ta giả sử)

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$$\Rightarrow$abc$\leq$1P=$\Sigma$$\frac{a^{2}b}{1+a+b}$$\leq$$\Sigma$$\frac{a^{2}b}{3\sqrt[3]{ab}}$=$\Sigma$$\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc$\leq$1$\Rightarrow$$a^{2}$$b^{2}$$c^{2}$$\Rightarrow$P$\leq$$\frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$$\Rightarrow$abc$\leq$1P=$\Sigma$$\frac{a^{2}b}{1+a+b}$$\leq$$\Sigma$$\frac{a^{2}b}{3\sqrt[3]{ab}}$=$\Sigma$$\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc$\leq$1$\Rightarrow$$a^{2}$$b^{2}$$c^{2}$$\leq$1$\Rightarrow $P$\leq$$\frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$$\Rightarrow$abc$\leq$1P=$\Sigma$$\frac{a^{2}b}{1+a+b}$$\leq$$\Sigma$$\frac{a^{2}b}{3\sqrt[3]{ab}}$=$\Sigma$$\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc$\leq$1$\Rightarrow$$a^{2}$$b^{2}$$c^{2}$$\Rightarrow$P$\leq$$\frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1

Ta có:a+b+c=3,a,b,c>0Áp dụng BĐT Cauchy:a+b+c$\geq$3$\sqrt[3]{abc}$$\Rightarrow$abc$\leq$1P=$\Sigma $$\frac{a^{2}b}{1+a+b}$$ \leq $$\Sigma$$ \frac{a^{2}b}{3\sqrt[3]{ab}}$=$\Sigma $$\frac{\sqrt[3]{a^{5}b^{2}}}{3}$Mà abc$\leq$1$\Rightarrow$$ a^{2} $$b^{2}$$ c^{2} $$\Rightarrow $P$\leq$$ \frac{a+b+c}{3}$=1(đpcm)Dấu''='' xra$\Leftrightarrow$a=b=c=1